Manipulating equations, also called solving equations or algebraic manipulation, refers to the process of performing various operations on mathematical equations to isolate a particular variable or find solutions that satisfy the given equation. This process is a fundamental aspect of algebra and is used extensively in mathematics, science, engineering, and various other fields.
The goal of manipulating equations is to simplify them, rearrange terms, and ultimately solve for the desired variable. The basic operations involved in manipulating equations include addition, subtraction, multiplication, division, and sometimes exponentiation.
Here's a general outline of how equation manipulation works
- Simplify Equations - Begin by simplifying both sides of the equation by combining like terms and performing arithmetic operations.
- Isolate the Variable - The objective is to isolate the variable you're interested in on one side of the equation. To do this, apply inverse operations to get rid of any constants or coefficients that are attached to the variable.
- Combine Like Terms - If there are any terms on the same side of the equation that contain the variable, combine them to simplify the equation.
- Use Inverse Operations - To isolate the variable, use inverse operations. For example, to undo addition, subtract the same value from both sides of the equation, to undo multiplication, divide both sides by the same value.
- Solve for the Variable - After isolating the variable, the equation should be in the form "variable = value" or "variable = expression." This provides the solution for the variable in terms of the other quantities in the equation.
Equation manipulation becomes more complex as equations involve higher order terms, variables on both sides, and various mathematical operations. It's an essential skill for solving problems in various disciplines, including physics, engineering, economics, and more. By skillfully manipulating equations, you can derive meaningful insights, make predictions, and solve real world problems.
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Manipulating Equations Rules
Manipulating equations involves performing various operations to transform an equation in order to simplify it, solve for a specific variable, or derive new relationships. Here are some general rules and principles to keep in mind when manipulating equations:
- Maintain equality - Any operation performed on one side of an equation must also be performed on the other side to maintain equality. This ensures that the equation remains balanced.
- Addition and subtraction - You can add or subtract the same quantity from both sides of an equation without changing its equality.
- Multiplication and division - You can multiply or divide both sides of an equation by the same non-zero value without changing its equality.
- Distributive property - When multiplying or dividing terms within parentheses, distribute the multiplication or division to each term inside the parentheses.
- Combining like terms - Combine terms that have the same variables raised to the same powers by adding or subtracting their coefficients.
- Factoring - Break down expressions into their constituent factors to simplify or solve equations. This is especially useful for solving quadratic equations.
- Substitution - Substitute one expression or variable with an equivalent expression to simplify or solve equations.
- Cross-multiplication - When dealing with fractions or ratios, you can cross-multiply by multiplying the numerator of one fraction by the denominator of another and vice versa.
- Taking logarithms or exponentials - Applying logarithmic or exponential functions to both sides of an equation can help simplify or solve equations involving exponents or logarithms.
- Rearranging terms - Change the order of terms within an equation to isolate variables or simplify expressions.
It's important to apply these rules carefully and consistently while manipulating equations to ensure that the equality of the original equation is maintained throughout the process. These are just a few techniques used to manipulate equations, and there are many more depending on the specific problem you're working on. It's important to maintain the equality of an equation by performing the same operation on both sides to ensure that the equation remains balanced throughout the manipulation process.
- The order of reareanging is important.
- B E D M A S Brackets Exponents Division Multiplication Addition Subtraction
- Solve -----> B E D M A S <----- Rearrange
- Use the opposite both sides.
- Every function has an opposite function.
- \(+ \; to \; -\)
- \(\times \; to \; \div\)
- \(x^2 \; to\; \sqrt{x} \)
- \(Sin \; to \;Sin^{-1}\)
- Whatever you do to one side of the equation you must do to the other side.
Manipulating Equation Examples
Addition, Subtraction and Divide Example
- Start with the equation: \(\large{2x - 3 = 7}\)
- To solve for \(\large{x}\) , we can follow these steps:
- Add \(\large{3}\) to both sides of the equation to isolate the term involving \(\large{x}\) : \(\large{2x - 3 + 3 = 7 + 3}\)
- Simplifying, we get: \(\large{2x = 10}\)
- Divide both sides of the equation by \(\large{2}\) to solve for \(\large{x}\) : \(\large{ \frac{2x}{2} = \frac{10}{2} }\)
- Simplifying, we get: \(\large{x = 5}\)
- Therefore, the solution for \(\large{x}\) is \(\large{x = 5}\)
Distributive property Example
- Start with the equation: \(\large{2 \left(x + 3 \right) = 10 }\)
- Apply the distributive property by multiplying \(\large{2}\) to both terms inside the parentheses: \(\large{2x + 2 \left(3\right) = 10}\)
- Simplifying, we get: \(\large{2x + 6 = 10}\)
- Subtract \(\large{6}\) from both sides of the equation to isolate the term involving \(\large{x}\): \(\large{2x + 6 - 6 = 10 - 6}\)
- Simplifying, we get: \(\large{2x = 4}\)
- Divide both sides of the equation by \(\large{2}\) to solve for \(\large{x}\): \(\large{ \frac{ 2x }{ 2 } = \frac{4}{2} }\)
- Simplifying, we get: \(\large{x = 2}\)
- Therefore, the solution for \(\large{x}\) is \(\large{x = 2}\)
Combining like terms Example
- Start with the equation: \(\large{3x + 2y + 5x - 4y = 12}\)
- Combine the terms with the same variable, \(\large{x}\) : \(\large{ (3x + 5x) + 2y - 4y = 12}\)
- Simplifying, we get: \(\large{8x + 2y - 4y = 12}\)
- Combine the terms with the same variable, \(\large{ y}\) : \(\large{8x + (2y - 4y) = 12}\)
- Simplifying, we get: \(\large{8x - 2y = 12}\)
- Therefore, the simplified equation is \(\large{8x - 2y = 12}\)
Factoring Example
- Start with the equation: \(\large{x^2 - 4x = 0}\)
- Observe that both terms on the left side have a common factor of \(\large{x}\) . Factor out \(\large{x}\) : \(\large{x(x - 4) = 0}\)
- Now, we have a product of two factors equal to zero. Set each factor equal to zero and solve for \(\large{x}\) : \(\large{x = 0}\) (from the factor \(\large{x = 0}\) ), \(\large{x - 4 = 0}\) (from the factor \(\large{x - 4 = 0}\) )
- Solving the second equation, we add \(\large{4}\) to both sides: \(\large{x = 4}\)
- Therefore, the solutions for \(\large{x}\) are \(\large{x = 0}\) and \(\large{x = 4}\)
Cross-multiplication Example
- Start with the equation: \(\large{\frac{3}{4} = \frac{x}{8} }\)
- Cross-multiply by multiplying the numerator of the left side with the denominator of the right side, and vice versa: \(\large{4 \left(x\right) = 3 \left(8\right) }\)
- Simplifying, we get: \(\large{4x = 24}\)
- Now, divide both sides of the equation by \(\large{4}\) to solve for \(\large{x}\) : \(\large{ \frac{4x}{ 4 } = \frac{24 }{ 4 } }\)
- Simplifying, we get: \(\large{x = 6}\)
- Therefore, the solution for \(\large{x}\) is \(\large{x = 6}\)
logarithms or exponentials Example
- Start with the equation: \(\large{2^{3x} = 16 }\)
- Take the logarithm of both sides of the equation, using a logarithm base of your choice (commonly, logarithm base 10 or natural logarithm base e): \(\large{log (base 2)(2^(3x)) = log (base 2)(16) }\)
- Applying the logarithmic property \(\large{log (base a)(a^b) = b}\) , we get: \(\large{3x * log (base 2)(2) = log (base 2)(16) }\)
- Simplifying, we get: \(\large{3x * 1 = log (base 2)(16) }\)
- Therefore, \(\large{3x = log (base 2)(16) }\)
- Now, divide both sides of the equation by \(\large{3}\) to solve for \(\large{x}\) : \(\large{ (3x) / 3 = log (base 2)(16) / 3 }\)
- Simplifying, we get: \(\large{x = log (base 2)(16) / 3 }\)
- Therefore, the solution for \(\large{x}\) is \(\large{x = log (base 2)(16) / 3 }\)
Canceling out a variable in a Formula
Canceling out a variable in a formula is a step often used in algebraic manipulations to simplify equations or isolate a specific variable. Here are some common scenarios in which canceling out variables is appropriate:
- Multiplication or Division by the Same Variable:
- If a variable appears in both the numerator and denominator of a fraction and is multiplied or divided by the same value, you can cancel it out. For example: \(\large{ \frac{2\;x}{x} = 2 }\) Here, the \(\large{x}\) in the numerator and the \(\large{x}\) in the denominator can be canceled out.
-
Adding or Subtracting the Same Variable on Both Sides:
- If the same variable is added or subtracted on both sides of an equation, you can cancel it out. For example: \(\large{ 2\;x + 5 = 3\;x +5 }\) Subtracting \(\large{5}\) from both sides cancels out the \(\large{5}\) terms.
-
Combining Like Terms:
- When simplifying expressions, you can cancel out like terms. For example: \(\large{ 3\;x + 2\;x = 5\;x }\) The \(\large{3\;x}\) and \(\large{2\;x}\) terms can be combined.
It's important to note that canceling out variables is a valid operation only when you are not dividing by zero, as division by zero is undefined in mathematics. Always be cautious and check the conditions under which the cancellation is valid. Additionally, if you're dealing with equations, make sure to apply the same operation to both sides to maintain the equality.
Manipulating Equation Arc - cosine, sine, tangent inverse
Arc - cosine, sine, tangent have an inverse
- This is okay:
- ArcCosine = \(cos^{-1}\)
- ArcSine=\(sin^{-1}\)
- ArcTan \(tan^{-1}\)
- This is not okay:
- \(cos^{-2}\), (notice the exponent is immediately to the right of the function, cos.
- \(sin^{-3}\)
- \(tan^{-2}\)
- but this is okay:
- \(cos{\theta}^{-2}\) Notice the exponent is on theta, so it would be cosine theta squared
- \((sin{\theta})^{-3}\) Notice the exponent is outside of the sin function. so this is also called sin theta sqared which is really confusing.
How does this work?
- If you want to solve for theta, you need to take the sin of both sides. This is a little bit different than what we've been normally doing but the equation then becomes - \(cos{\frac{O}{H}}= \theta\)
- As an example, for the equation - \(\frac{O}{H} = cos^{-1} {\frac{\theta}{\rho}}\)
- To solve for theta, take the cos of both sides - \(cos{\frac{O}{H}} = \frac{\theta}{\rho}\)
- Then multiply both sides by rho - \(\rho cos{\frac{O}{H}} = \theta\)
Manipulating Equation Formulas
Manipulate Equation 1 \(\large{ a = b \; c }\)
- Solve for variable \(\large{b}\) in the equation \(\large{a = b \; c}\) .
- Divide both sides of the equation by \(\large{c}\) : \(\large{ \frac{a}{c} = \frac{b \; c}{ c} }\)
- Simplify the right-hand side: \(\large{ \frac{a}{c} = b \; \frac{c}{c} }\)
- Since \(\large{ \frac{c}{c} }\) is equal to \(\large{1}\) , the equation becomes: \(\large{ \frac{a}{c} = b \; 1 }\)
- \(\large{b}\) is isolated on one side of the equation. The solution for \(\large{b}\) is: \(\large{ b = \frac{a}{c} }\)
- Solve for variable \(\large{c}\) in the equation \(\large{a = b \; c}\) .
- Divide both sides of the equation by \(\large{b}\) : \(\large{ \frac{a}{b} = \frac{b \; c}{ b} }\)
- Simplify the right-hand side: a / b = c * (b / b) \(\large{ \frac{a}{b} = c \; \frac{b}{b} }\)
- Since \(\large{ \frac{b}{b} }\) is equal to \(\large{1}\) , the equation becomes: \(\large{ \frac{a}{b} = c \; 1 }\)
- \(\large{c}\) is isolated on one side of the equation. The solution for \(\large{c}\) is: \(\large{ c = \frac{a}{b} }\)
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formulas | Formulas | Formulas | Formulas | Formulas |
|
\(\large{ a = b \; c }\) \(\large{ b = \frac { a }{ c } }\) \(\large{ c = \frac { a }{ b } }\) |
\(\large{ a = b \; c \; d }\) \(\large{ b = \frac{ a }{ c \; d } }\) \(\large{ c = \frac{ a }{ b \; d } }\) \(\large{ d = \frac{ a }{ b \; c } }\) |
\(\large{ a = b \; c \; d \; e }\) \(\large{ b = \frac{ a }{ c \; d \; e } }\) \(\large{ c = \frac{ a }{ b \; d \; e } }\) \(\large{ d = \frac{ a }{ b \; c \; e } }\) \(\large{ e = \frac{ a }{ b \; c \; d } }\) |
\(\large{ a = b \; c \; d \; e \; f }\) \(\large{ b = \frac{ a }{ c \; d \; e \; f } }\) \(\large{ c = \frac{ a }{ b \; d \; e \; f } }\) \(\large{ d = \frac{ a }{ b \; c \; e \; f } }\) \(\large{ e = \frac{ a }{ b \; c \; d \; f } }\) \(\large{ f = \frac{ a }{ b \; c \; d \; e } }\) |
\(\large{ a = b \; c \; d \; e \; f \; g }\) \(\large{ b = \frac{ a }{ c \; d \; e \; f \; g} }\) \(\large{ c = \frac{ a }{ b \; d \; e \; f \; g} }\) \(\large{ d = \frac{ a }{ b \; c \; e \; f \; g } }\) \(\large{ e = \frac{ a }{ b \; c \; d \; f \; g} }\) \(\large{ f = \frac{ a }{ b \; c \; d \; e \; g } }\) \(\large{ g = \frac{ a }{ b \; c \; d \; e \; f } }\) |
Manipulate Equation 2 \(\large{ a = \frac{ b }{c} }\)
- Solve for variable \(\large{b}\) in the equation \(\large{ a = \frac{ b }{c} }\) .
-
- Multiply both sides of the equation by \(\large{c}\) : \(\large{a \; c = b}\)
- \(\large{b}\) is isolated on one side of the equation. The solution for \(\large{b}\) is: \(\large{b = a \; c }\)
-
Solve for variable \(\large{c}\) in the equation \(\large{ a = \frac{ b }{c} }\) .
-
- Multiply both sides of the equation by \(\large{c}\) : \(\large{a \; c = b}\)
- Now, divide both sides of the equation by \(\large{a}\) : \(\large{c = \frac{b}{a} }\)
- \(\large{c}\) is isolated on one side of the equation. The solution for \(\large{c}\) is: \(\large{c = \frac{b}{a} }\)
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formulas | Formulas | Formulas | Formulas | Formulas |
|
\(\large{ a = \frac{ b }{c} }\) \(\large{ b = a\;c }\) \(\large{ c = \frac{ b }{a} }\) |
\(\large{ a = \frac{ b\;c }{d} }\) \(\large{ b = \frac{ a\;d }{c} }\) \(\large{ c = \frac{ a\;d }{b} }\) \(\large{ d = \frac{ b\;c }{a} }\) |
\(\large{ a = \frac{ b\;c\;d }{e} }\) \(\large{ b = \frac{ a\;e }{c\;d} }\) \(\large{ c = \frac{ a\;e }{b\;d} }\) \(\large{ d = \frac{ a\;e }{b\;c} }\) \(\large{ e = \frac{ b\;c\;d }{a} }\) |
\(\large{ a = \frac{ b\;c\;d\;e }{ f } }\) \(\large{ b = \frac{ a\;f }{ c\;d\;e } }\) \(\large{ c = \frac{ a\;f }{ b\;d\;e } }\) \(\large{ d = \frac{ a\;f }{ b\;c\;e } }\) \(\large{ e = \frac{ a\;f }{ b\;c\;d } }\) \(\large{ f = \frac{ b\;c\;d\;e }{ a } }\) |
\(\large{ a = \frac{ b\;c\;d\;e\;f }{ g } }\) \(\large{ b = \frac{ a\;g }{ c\;d\;e\;f } }\) \(\large{ c = \frac{ a\;g }{ b\;d\;e\;f } }\) \(\large{ d = \frac{ a\;g }{ b\;c\;e\;f } }\) \(\large{ e = \frac{ a\;g }{ b\;c\;d\;f } }\) \(\large{ f = \frac{ a\;g }{ b\;c\;d\;e } }\) \(\large{ g = \frac{ b\;c\;d\;e\;f }{ a } }\) |
Manipulate Equation 3 \(\large{ a = b + c }\)
- Solve for variable \(\large{b}\) in the equation \(\large{a = b + c}\) .
- Subtract \(\large{c}\) from both sides of the equation: \(\large{a - c = b + c - c}\)
- Simplify the right-hand side: \(\large{a - c = b \; \left(1 \right) }\)
- Since \(\large{b \; 1}\) is equal to \(\large{b}\) , the equation becomes: \(\large{a - c = b}\)
- \(\large{b}\) is isolated on one side of the equation. The solution for \(\large{b}\) is: \(\large{b = a - c}\)
- Solve for variable \(\large{c}\) in the equation \(\large{a = b + c}\) .
- Subtract \(\large{b}\) from both sides of the equation: \(\large{a - b = b + c - b}\)
- Simplify the right-hand side: \(\large{a - b = c \; \left(1 \right) }\)
- Since \(\large{c \; 1}\) is equal to \(\large{c}\) , the equation becomes: \(\large{a - b = c}\)
- \(\large{c}\) is isolated on one side of the equation. The solution for \(\large{c}\) is: \(\large{c = a - b}\)
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
|
\(\large{ a = b + c }\) \(\large{ b = a - c }\) \(\large{ c = a - b }\) |
\(\large{ a = b + c\;d }\) \(\large{ b = a - c\;d }\) \(\large{ c = \frac{ a\;-\;b }{ d } }\) \(\large{ d = \frac{ a\;-\;b }{ c } }\) |
\(\large{ a = b + c\;d\;e }\) \(\large{ b = a - c\;d\;e }\) \(\large{ c = \frac{ a\;-\;b }{ d\;e } }\) \(\large{ d = \frac{ a\;-\;b }{ c\;e } }\) \(\large{ e = \frac{ a\;-\;b }{ c\;d } }\) |
||
Manipulate Equation 4 \(\large{ a = \frac{ b^2 }{ c } }\)
- Solve for variable \(\large{b}\) in the equation \(\large{ a = \frac{ b^2 }{ c } }\) .
- Multiply both sides of the equation by \(\large{c}\) : \(\large{a \; c = b^2}\)
- Take the square root of both sides to isolate \(\large{b}\) : \(\large{b = \sqrt{a \;c } }\)
- However, when taking the square root, we need to consider both the positive and negative square roots since \(\large{b^2}\) can be positive or negative. \(\large{b = ± \sqrt{a \;c } }\)
- Solve for variable \(\large{c}\) in the equation \(\large{ a = \frac{ b^2 }{ c } }\) .
- Multiply both sides of the equation by \(\large{c}\) : \(\large{a \; c = b^2 }\)
- Divide both sides of the equation by \(\large{a}\) : \(\large{c = \frac{b^2 }{ a } }\)
- The solution for \(\large{c}\) is \(\large{c = \frac{b^2 }{ a} }\)
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
|
\(\large{ a = \frac{ b^2 }{ c } }\) \(\large{ b = \sqrt{ a \; c } }\) \(\large{ c = \frac{ b^2 }{ a } }\) |
\(\large{ a = \frac{ b^2\;c }{ d } }\) \(\large{ b = \sqrt{ \frac{ a \; d }{ c } } }\) \(\large{ c = \frac{ a \; d }{ b^2 } }\) \(\large{ d = \frac{ b^2\;c }{ a } }\) |
\(\large{ a = \frac{ b^2\;c\;d }{e} }\) \(\large{ b = \sqrt{ \frac{ a \; e }{ c \; d } } }\) \(\large{ c = \frac{ a \; e }{ b^2 \; d } }\) \(\large{ d = \frac{ a \; e }{ b^2 \; c } }\) \(\large{ e = \frac{ b^2\;c\;d }{ a } }\) |
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Manipulate Equation 5 \(\large{a = \sqrt{\frac{b}{c} } }\)
- Solve for variable \(\large{b}\) in the equation \(\large{a = \sqrt{\frac{b}{c} } }\) .
- Square both sides of the equation to eliminate the square root: \(\large{a^2 = \frac{b }{ c} }\)
- Multiply both sides of the equation by \(\large{c}\) to get rid of the fraction: \(\large{a^2 \;c = b}\)
- \(\large{b}\) is isolated on one side of the equation. The solution for \(\large{b}\) is \(\large{b = a^2 \; c }\) .
- Solve for variable \(\large{c}\) in the equation \(\large{a = \sqrt{\frac{b}{c} } }\) .
- Square both sides of the equation to eliminate the square root: \(\large{a^2 = \frac{b }{ c } }\)
- Multiply both sides of the equation by \(\large{c}\) to isolate \(\large{c}\) on one side: \(\large{c \; a^2 = b}\)
- Divide both sides of the equation by \(\large{a^2}\) to solve for \(\large{c}\) : \(\large{c = \frac{b }{ a^2} }\)
- The solution for \(\large{c}\) is \(\large{c = \frac{b }{ a^2} }\) .
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
|
\(\large{a = \sqrt{\frac{b}{c} } }\) \(\large{b = a^2 \; c }\) \(\large{c = \frac{b}{a^2} }\) |
\(\large{a = \sqrt{\frac{b\;c}{d} } }\) \(\large{b = \frac{a^2\;d}{c} }\) \(\large{c = \frac{a^2\;d}{b} }\) \(\large{d = \frac{b\;c}{a^2} }\) |
\(\large{a = \sqrt{\frac{b\;c\;d}{e} } }\) \(\large{b = \frac{a^2\;e}{c\;d} }\) \(\large{c = \frac{a^2\;e}{b\;d} }\) \(\large{d = \frac{a^2\;e}{b\;c} }\) \(\large{e = \sqrt{\frac{b\;c\;d}{a^2} } }\) |
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Manipulate Equation 6 \(\large{ a \; b = c \; d }\)
- Solve for variable \(\large{a}\) in the equation \(\large{ a \; b = c \; d }\) .
- Divide both sides of the equation by \(\large{b}\) : \(\large{ \frac{a \; b}{ b} = \frac{c \; d}{ b} }\)
- Simplify the left side of the equation \(\large{ \frac{b}{b} = 1}\) : \(\large{b = \frac{c \; d}{ b} }\)
- The solution for \(\large{a}\) is: \(\large{ a = \frac{ c \;d }{b} }\)
- Solve for variable \(\large{b}\) in the equation \(\large{ a \; b = c \; d }\) .
- Divide both sides of the equation by \(\large{a}\) : \(\large{ \frac{a \; b}{ a} = \frac{c \; d}{ a} }\)
- Simplify the left side of the equation \(\large{ \frac{a}{a} = 1}\) : \(\large{a = \frac{c \; d}{ a} }\)
- The solution for \(\large{b}\) is: \(\large{ b = \frac{ c \;d }{a} }\)
- Solve for variable \(\large{c}\) in the equation \(\large{ a \; b = c \; d }\) .
- Divide both sides of the equation by \(\large{d}\) : \(\large{ \frac{a \; b}{ d} = \frac{c \; d}{ d} }\)
- Simplify the left side of the equation \(\large{ \frac{d}{d} = 1}\) : \(\large{ \frac{a \; b}{ d} = c }\)
- \(\large{c}\) is isolated on one side of the equation. The solution for \(\large{c}\) is: \(\large{ c = \frac{ a \;b }{d} }\)
- Solve for variable \(\large{d}\) in the equation \(\large{ a \; b = c \; d }\) .
- Divide both sides of the equation by \(\large{d}\) : \(\large{ \frac{a \; b}{ c} = \frac{c \; d}{ c} }\)
- Simplify the left side of the equation \(\large{ \frac{c}{c} = 1}\) : \(\large{ \frac{a \; b}{ c} = d }\)
- \(\large{d}\) is isolated on one side of the equation. The solution for \(\large{d}\) is: \(\large{ d = \frac{ a \;b }{c} }\)
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
|
\(\large{ a \; b = c \; d }\) \(\large{ a = \frac{ c \;d }{b} }\) \(\large{ b = \frac{ c \;d }{a} }\) \(\large{ c = \frac{ a \;b }{d} }\) \(\large{ d = \frac{ a \;b }{c} }\) |
\(\large{ a \; b = c\;d\;e }\) \(\large{ a = \frac{ c\;d\;e }{b} }\) \(\large{ b = \frac{ c\;d\;e }{a} }\) \(\large{ c = \frac{ a \;b }{d\;e} }\) \(\large{ d = \frac{ a \;b }{c\;e} }\) \(\large{ e = \frac{ a \;b }{c\;d} }\) |
\(\large{ a \; b = c\;d\;e\;f }\) \(\large{ a = \frac{ c\;d\;e\;f }{b} }\) \(\large{ b = \frac{ c\;d\;e\;f }{a} }\) \(\large{ c = \frac{ a\;b }{d\;e\;f} }\) \(\large{ d = \frac{ a\;b }{c\;e\;f} }\) \(\large{ e = \frac{ a\;b }{c\;d\;f} }\) \(\large{ f = \frac{ a\;b }{c\;d\;e} }\) |
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Manipulate Equation 7 \(\large{ a = b \; \frac{ c }{ d } }\)
- Solve for variable \(\large{b}\) in the equation \(\large{ a = b \; \frac{ c }{ d } }\).
- Multiply both sides of the equation by \(\large{d}\) to eliminate the fraction: \(\large{ a\;d = b\;c }\)
- Divide both sides of the equation by \(\large{c}\) to isolate \(\large{b}\) : \(\large{ b = \frac{a\;d}{c} }\)
- The solution for \(\large{b}\) is \(\large{ b = \frac{a\;d}{c} }\) .
- Multiply both sides of the equation by \(\large{d}\) to eliminate the fraction: \(\large{ a\;d = b\;c }\)
- Solve for variable \(\large{c}\) in the equation \(\large{ a = b \; \frac{ c }{ d } }\)
- Multiply both sides of the equation by \(\large{d}\) to eliminate the fraction: \(\large{ a\;d = b\;c }\)
- Divide both sides of the equation by \(\large{b}\) to isolate \(\large{c}\) : \(\large{ c = \frac{a\;d}{b} }\)
- The solution for \(\large{c}\) is \(\large{ c = \frac{a\;d}{b} }\) .
- Solve for variable \(\large{d}\) in the equation \(\large{ a = b \; \frac{ c }{ d } }\)
- Multiply both sides of the equation by \(\large{d}\) to eliminate the fraction: \(\large{ a\;d = b\;c }\)
- Divide both sides of the equation by \(\large{a}\) to isolate \(\large{d}\) : \(\large{ d = \frac{b\;c}{a} }\)
- The solution for \(\large{d}\) is \(\large{ d = \frac{b\;c}{a} }\) .
| Manipulating Equations | ||||
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| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = b \; \frac{ c }{ d } }\) \(\large{ b = \frac{ a \; d }{ c } }\) \(\large{ c = \frac{ a \; d }{ b } }\) \(\large{ d = \frac{ b \; c }{ a } }\) |
\(\large{ a = b \; c \; \frac{ d }{ e } }\) \(\large{ b = \frac{ a\;e }{ c\;d } }\) \(\large{ c = \frac{ a\;e }{ b\;d } }\) \(\large{ d = \frac{ a\;e }{ b\;c } }\) \(\large{ e = \frac{ b\;c\;d }{ a } }\) |
\(\large{ a = b\;c\;d \frac{e}{f} }\) \(\large{ b = \frac{ a }{c\;d\;\frac{e}{f} } }\) \(\large{ c = \frac{ a }{b\;d\;\frac{e}{f} } }\) \(\large{ d = \frac{ a }{b\;c\;\frac{e}{f} } }\) \(\large{ e = \frac{ a\;f }{b\;c\;d} }\) \(\large{ f = \frac{ b\;c\;d\;e }{a} }\) |
\(\large{ a = \frac{ b }{ c } \; d}\) \(\large{ b = \frac{ a \; c }{ d } }\) \(\large{ c = \frac{ b \; d }{ a } }\) \(\large{ d = \frac{ a \; c }{ b } }\) |
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Manipulate Equation 8 \(\large{ a = b\; c^2 }\)
- Solve for variable \(\large{b}\) in the equation \(\large{ a = b\; c^2 }\).
- Divide both sides of the equation by \(\large{c^2}\) : \(\large{ \frac{a }{ c^2} = b \; \frac{c^2 }{ c^2 } }\)
- Simplify the right side of the equation: \(\large{ \frac{a }{ c^2} = b \; 1 }\)
- Divide both sides of the equation by \(\large{1}\) (which doesn't change the equation): \(\large{ \frac{a }{ c^2} = b }\)
- \(\large{b}\) is isolated on one side of the equation. The solution for \(\large{b}\) is: \(\large{ b = \frac{a }{ c^2} }\)
- Solve for variable \(\large{c}\) in the equation \(\large{ a = b\; c^2 }\).
- Divide both sides of the equation by \(\large{b}\) : \(\large{ \frac{a }{ b} = c^2 }\)
- Take the square root of both sides of the equation: \(\large{ \sqrt{ \frac{a }{ b} } = \sqrt{ c^2 } }\)
- Simplify the right side of the equation: \(\large{ \sqrt{ \frac{a }{ b} } = c }\)
- \(\large{c}\) is isolated on one side of the equation. The solution for \(\large{c}\) is: \(\large{ c = \sqrt{ \frac{a }{ b} } }\)
| Manipulating Equations | ||||
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| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = b\; c^2 }\) \(\large{ b = \frac{a}{c^2} }\) \(\large{ c = \sqrt{ \frac{a}{b} } }\) |
\(\large{ a = b\; c^3 }\) \(\large{ b = \frac{a}{c^3} }\) \(\large{ c = \frac{a}{b}^{\frac{1}{3}} }\) |
\(\large{ a = b\; c^4 }\) \(\large{ b = \frac{a}{c^4} }\) \(\large{ c = \frac{a}{b}^{\frac{1}{4}} }\) |
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Manipulate Equation 9 \(\large{ a = \frac{1}{2} \;b}\)
- Solve for variable \(\large{b}\) in the equation .
- Multiply both sides of the equation by \(\large{2}\) (to get rid of the fraction): \(\large{ 2 \; a = 2 \; \frac{1}{2} \; b }\)
- Simplify the right side of the equation: \(\large{ 2 \; a = b}\)
- \(\large{b}\) is isolated on one side of the equation. So, the solution for \(\large{b}\) is: \(\large{ b = 2\;a }\)
| Manipulating Equations | ||||
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| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = \frac{1}{2} \;b}\) \(\large{ b = 2\;a }\) |
\(\large{ a = \frac{1}{2} \;b\;c }\) \(\large{ b = \frac{2\;a}{c} }\) \(\large{ c = \frac{2\;a}{b} }\) |
\(\large{ a = \frac{1}{2} \;b\;c\;d }\) \(\large{ b = \frac{2\;a}{c\;d} }\) \(\large{ c = \frac{2\;a}{b\;d} }\) \(\large{ d = \frac{2\;a}{b\;c} }\) |
\(\large{ a = \frac{1}{2} \;b\;c\;d\;e }\) \(\large{ b = \frac{2\;a}{c\;d\;e} }\) \(\large{ c = \frac{2\;a}{b\;d\;e} }\) \(\large{ d = \frac{2\;a}{b\;c\;e} }\) \(\large{ e = \frac{2\;a}{b\;c\;d} }\) |
\(\large{ a = \frac{1}{2} \;b\;c\;d\;e\;f }\) \(\large{ b = \frac{2\;a}{c\;d\;e\;f} }\) \(\large{ c = \frac{2\;a}{b\;d\;e\;f} }\) \(\large{ d = \frac{2\;a}{b\;c\;e\;f} }\) \(\large{ e = \frac{2\;a}{b\;c\;d\;f} }\) \(\large{ f = \frac{2\;a}{b\;c\;d\;e} }\) |
Manipulate Equation 10 - \(\large{ a = \frac{1}{2} \; b + c }\)
- Solve for variable \(\large{b}\) in the equation \(\large{ a = \frac{1}{2} \; b + c }\).
- Subtract both sides by \(\large{c}\): \(\large{ a - c = \frac{1}{2} \; b }\).
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = \frac{1}{2} \; b + c }\) \(\large{ b = 2\;a - c }\) \(\large{ c = 2\;a - b }\) |
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Manipulate Equation 11 - \(\large{ a = \frac{1}{2} \; b \; c^2 }\)
- Solve for variable \(\large{b}\) in the equation \(\large{ a = \frac{1}{2} \; b \; c^2 }\).
- Multiply both sides by \(\large{2}\) to get rid of the fraction: \(\large{ 2 \; a = b \; c^2 }\).
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = \frac{1}{2} \; b^2 }\) \(\large{ c = \sqrt{ 2 \; a } }\) |
\(\large{ a = \frac{1}{2} \; b \; c^2 }\) \(\large{ b = \frac{ 2\;a }{ c^2 } }\) \(\large{ c = \sqrt{ \frac{ 2\;a }{ b } } }\) |
\(\large{ a = \frac{1}{2} \; b \; c \; d^2 }\) \(\large{ b = \frac{ 2\;a }{ c \; d^2 } }\) \(\large{ c = \frac{ 2\;a }{ b \; d^2 } }\) \(\large{ c = \sqrt{ \frac{ 2\;a }{ b \; c } } }\) |
\(\large{ a = \frac{1}{2} \; b \; c \; d \; e^2 }\) \(\large{ b = \frac{ 2\;a }{ c \; d \; e^2 } }\) \(\large{ c = \frac{ 2\;a }{ b \; d \; e^2 } }\) \(\large{ d = \frac{ 2\;a }{ b \; c \; e^2 } }\) \(\large{ c = \sqrt{ \frac{ 2\;a }{ b \; c \; d } } }\) |
\(\large{ a = \frac{1}{2} \; b \; c \; d \; e \; f^2 }\) \(\large{ b = \frac{ 2\;a }{ c \; d \; e \; f^2 } }\) \(\large{ c = \frac{ 2\;a }{ b \; d \; e \; f^2 } }\) \(\large{ d = \frac{ 2\;a }{ b \; c \; e \; f^2 } }\) \(\large{ e = \frac{ 2\;a }{ b \; c \; d \; f^2 } }\) \(\large{ f = \sqrt{ \frac{ 2\;a }{ b \; c \; d \; e} } }\) |
Manipulate Equation 12 - \(\large{ a = \frac{ b \; \left( c \;-\; d \right) }{ e } }\)
- Solve for variable \(\large{b}\) in the equation \(\large{ a = \frac{ b \; \left( c \;-\; d \right) }{ e } }\) .
- Multiply both sides by \(\large{e}\) to get rid of the fraction: \(\large{ a \; e = b \; \left( c \;-\; d \right) }\).
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = \frac{ b \; \left( c \;-\; d \right) }{ e } }\) \(\large{ b = \frac{ a \; e }{ c \;-\; d } }\) \(\large{ c = \frac{ a \; e }{ b } + d }\) \(\large{ d = c - \frac{ a \; e }{ b } }\) \(\large{ e = \frac{ b \; \left( c \;-\; d \right) }{ a } }\) |
\(\large{ a = \frac{ b \; c \; \left( d \;-\; e \right) }{ f } }\) \(\large{ b = \frac{ a \; f }{ c \; \left( d \;-\; e \right) } }\) \(\large{ c = \frac{ a \; f }{ b \; \left( d \;-\; e \right) } }\) \(\large{ d = \frac{ a \; f }{ b \; c } + e }\) \(\large{ e = d - \frac{ a \; f }{ b \; c } }\) \(\large{ f = \frac{ b \; c \; \left( d \;-\; e \right) }{ a } }\) |
\(\large{ a = \frac{ b \; c \; d \; \left( e \;-\; f \right) }{ g } }\) \(\large{ b = \frac{ a \; g }{ c \; d \; \left( e \;-\; f \right) } }\) \(\large{ c = \frac{ a \; g }{ b \; d \; \left( e \;-\; f \right) } }\) \(\large{ d = \frac{ a \; g }{ b \; c \; \left( e \;-\; f \right) } }\) \(\large{ e = \frac{ a \; g }{ b \; c \; d } + f }\) \(\large{ f = e \;-\; \frac{ a \; g }{ b \; c \; d } }\) \(\large{ g = \frac{ b \; c \; d \; \left( e \;-\; f \right) }{ a } }\) |
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Manipulate Equation 13 - Miscellaneous
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = \frac{ 2\; \left( b \;-\; c \right) }{d\;e^2} }\) \(\large{ b = \frac{ a\;d\;e^2 }{ 2 } \;+ c }\) \(\large{ c = b -\; \frac{ a\;d\;e^2 }{ 2 } }\) \(\large{ d = \frac{ 2\; \left( b \;-\; c \right) }{a\;e^2} }\) \(\large{ e = \sqrt{ \frac{ 2\; \left( b \;-\; c \right) }{a\;d} } }\) |
\(\large{ a = b\; \sqrt{ c\; d } }\) \(\large{ b = \frac{a}{ \sqrt{ c\; d } } }\) \(\large{ c = \frac{1}{d} \; \left( \frac{a}{b} \right)^2 }\) \(\large{ d = \frac{1}{c} \; \left( \frac{a}{b} \right)^2 }\) |
\(\large{ a = \frac{1}{b} \; \left( c \right)^{\frac{1}{6}} }\) \(\large{ b = \frac{1}{a} \; \left( c \right)^{\frac{1}{6}} }\) \(\large{ c = \left( a\;b \right)^6 }\) |
\(\large{ a = \frac{ b\;c\;d^2 }{ 2\;e\;f } }\) \(\large{ b = \frac{2\; a\;e\;f }{ c\;d^2 } }\) \(\large{ c = \frac{2\; a\;e\;f }{ b\;d^2 } }\) \(\large{ d = \sqrt{ \frac{2\; a\;e\;f }{ b\;c } } }\) \(\large{ e = \frac{ b\;c\;d^2 }{ 2\;a\;f } }\) \(\large{ f = \frac{ b\;c\;d^2 }{ 2\;a\;e } }\) |
\(\large{ a = - b \left( c - d \right) }\) \(\large{ b = \frac{ a }{ c \;-\; d } }\) \(\large{ c = d \;- \; \frac{ a }{ b } }\) \(\large{ d = \frac{ a }{ b } + c }\) |
Manipulate Equation 14 - Miscellaneous
| Manipulating Equations | ||||
|---|---|---|---|---|
| Formula | Formula | Formula | Formula | Formula |
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\(\large{ \frac{ a }{ b } = \frac{ c }{ d } }\) \(\large{ a = \frac{ b \; c }{ d } }\) \(\large{ b = \frac{ a \; d }{ c } }\) \(\large{ c = \frac{ a \; d }{ b } }\) \(\large{ d = \frac{ b \; c }{ a } }\) |
\(\large{ \frac{ a \; b }{ c } = \frac{ d \; e }{ f } }\) \(\large{ a = \frac{ c\;d\;e }{ b\;f } }\) \(\large{ b = \frac{ c\;d\;e }{ a\;f } }\) \(\large{ c = \frac{ a\;b\;f }{ d\;e } }\) \(\large{ d = \frac{ a\;b\;f }{ c\;e } }\) \(\large{ e = \frac{ a\;b\;f }{ c\;d } }\) \(\large{ f = \frac{ c\;d\;e }{ a\;b } }\) |
\(\large{ a = \frac{ b \; c \; d }{ e \; f } }\) \(\large{ b = \frac{ a \; e \; f }{ c \; d } }\) \(\large{ c = \frac{ a \; e \; f }{ b \; d } }\) \(\large{ d = \frac{ a \; e \; f }{ b \; c } }\) \(\large{ e = \frac{ b \; c \; d }{ a \; f } }\) \(\large{ f = \frac{ b \; c \; d }{ a \; e } }\) |
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Manipulate Equation 15 - Miscellaneous
| Manipulating Equations | ||||
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| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = b - \frac{ c }{ d } }\) \(\large{ b = a + \frac{ c }{ d } }\) \(\large{ c = b\;d - a\;d }\) \(\large{ d = \frac{ c }{ b\;-\; a } }\) |
\(\large{ a = \frac{ b }{ c } + d }\) \(\large{ b = c \; \left(a - d \right) }\) \(\large{ c = \frac{ b }{ a \;-\;d } }\) \(\large{ d = a - \frac{ b }{ c } }\) |
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Manipulate Equation 16 - \(\large{ a = \frac{ b \;+\; c }{ d } }\)
| Manipulating Equations | ||||
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| Formula | Formula | Formula | Formula | Formula |
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\(\large{ a = \frac{ b \;+\; c }{ d } }\) \(\large{ b = a - \frac{ c }{ d } }\) \(\large{ c = a \; d - b \; d }\) \(\large{ d = \frac{ c }{ a \;-\; b } }\) |
\(\large{ a = \frac{ b \;-\; c }{ d } }\) \(\large{ b = a \;c - a \; d }\) \(\large{ c = \frac{ a \; d \;+\; b }{ a } }\) \(\large{ d = \frac{ a \; c \;-\; b }{ a } }\) |
\(\large{ a = \frac{ b }{ c \;+\; d } }\) \(\large{ b = a \; c + a \; d }\) \(\large{ c = \frac{ b \;-\; a \; d }{ a } }\) \(\large{ d = \frac{ b \;-\; a \; c }{ a } }\) |
\(\large{ a = \frac{ b }{ c \;-\; d } }\) \(\large{ b = a \; c - a \; d }\) \(\large{ c = \frac{ a \; d \;+\; b }{ a } }\) \(\large{ d = \frac{ a \; c \;-\; b }{ a } }\) |
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