Rearrange Formulas and Solve Equations: CCSS.Math.Content.HSA-CED.A.4 - Common Core: High School - Algebra

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Question

To calculate the slope of a line, the following formula is used.

$m=\frac{y_2-y_1}{x_2-x_1}$

Solve this equation for .

Answer

This question is asking to rearrange the slope formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

First, multiply both sides by the quantity in order to move it from the right side of the equation to the left side. Performing the opposite operation allows for rearranging to occur.

$\(x_2-x_1)\cdot m=\frac{y_2-y_1}{x_2-x_1}\cdot (x_2-x_1) \\(x_2-x_1)m=y_2-y_1$

Next, add to both sides of the equation.

$\(x_2-x_1)m+y_1=y_2-y_1+y_1 \(x_2-x_1)m+y_1=y_2$

Therefore, the equation solved for is

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Question

To calculate the slope of a line, the following formula is used.

$m=\frac{y_2-y_1}{x_2-x_1}$

Solve this equation for .

Answer

This question is asking to rearrange the slope formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

First, multiply both sides by the quantity in order to move it from the right side of the equation to the left side. Performing the opposite operation allows for rearranging to occur.

$\(x_2-x_1)\cdot m=\frac{y_2-y_1}{x_2-x_1}\cdot (x_2-x_1) \\(x_2-x_1)m=y_2-y_1$

Next, subtract to both sides of the equation.

$\(x_2-x_1)m-y_2=y_2-y_1-y_2 \(x_2-x_1)m-y_2=-y_1$

Now, divide by negative one on both sides.

$\frac{(x_2-x_1)m-y_2}{-1}=-\frac{y_1}{-1}$

Therefore, the equation solved for is,

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Question

To calculate the slope of a line, the following formula is used.

$m=\frac{y_2-y_1}{x_2-x_1}$

Solve this equation for

Answer

This question is asking to rearrange the slope formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

First, multiply both sides by the quantity in order to move it from the right side of the equation to the left side. Performing the opposite operation allows for rearranging to occur.

$\(x_2-x_1)\cdot m=\frac{y_2-y_1}{x_2-x_1}\cdot (x_2-x_1) \\(x_2-x_1)m=y_2-y_1$

Next divide by on both sides.

$\frac{(x_2-x_1)m}{m}=\frac{y_2-y_1}{m}$

$x_2-x_1=\frac{y_2-y_1}{m}$

Now add to both sides.

$x_2-x_1+x_1=\frac{y_2-y_1}{m}+x_1$

$x_2=\frac{y_2-y_1}{m}+x_1$

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Question

To calculate the slope of a line, the following formula is used.

$m=\frac{y_2-y_1}{x_2-x_1}$

Solve this equation for .

Answer

This question is asking to rearrange the slope formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

First, multiply both sides by the quantity in order to move it from the right side of the equation to the left side. Performing the opposite operation allows for rearranging to occur.

$\(x_2-x_1)\cdot m=\frac{y_2-y_1}{x_2-x_1}\cdot (x_2-x_1) \\(x_2-x_1)m=y_2-y_1$

Next divide by on both sides.

$\frac{(x_2-x_1)m}{m}=\frac{y_2-y_1}{m}$

$x_2-x_1=\frac{y_2-y_1}{m}$

Now subtract from both sides.

$x_2-x_1-x_2=\frac{y_2-y_1}{m}-x_2$

$-x_1=\frac{y_2-y_1}{m}-x_2$

Finally, divide by negative one.

$x_1=-\frac{y_2-y_1}{m}+x_2$

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Question

The equation for a circle centered at (0,0) is,

Solve the equation for .

Answer

This question is asking to rearrange this formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

To isolate take the square root of each side.

$\sqrt{x^2+y^2}=\sqrt{r^2}$

$\sqrt{x^2+y^2}=r$

Therefore, the equation solved for is

$r=\sqrt{x^2+y^2}$

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Question

The equation for a circle centered at (0,0) is,

Solve the equation for .

Answer

This question is asking to rearrange this formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

To isolate subtract from both sides and then take the square root.

$\sqrt{x^2}=\sqrt{r^2-y^2}$

Therefore, the equation solved for is

$x=\sqrt{r^2-y^2}$

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Question

The equation for a circle centered at (0,0) is,

Solve the equation for .

Answer

This question is asking to rearrange this formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

To isolate subtract from both sides and then take the square root.

$\sqrt{y^2}=\sqrt{r^2-x^2}$

Therefore, the equation solved for is

$y=\sqrt{r^2-x^2}$

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Question

To calculate simple interest the following formula is used.

Solve the equation for .

Answer

This question is asking to rearrange this formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since is being multiplied by , divide by on both sides.

$\frac{I}{RT}=\frac{PRT}{RT}$

$\frac{I}{RT}=P$

Therefore, the equation solved for is,

$P=\frac{I}{RT}$

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Question

To calculate simple interest the following formula is used.

Solve the equation for .

Answer

This question is asking to rearrange this formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since is being multiplied by , divide by on both sides.

$\frac{I}{PT}=\frac{PRT}{PT}$

$\frac{I}{PT}=R$

Therefore, the equation solved for is,

$R=\frac{I}{PT}$

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Question

To calculate simple interest the following formula is used.

Solve the equation for .

Answer

This question is asking to rearrange this formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since is being multiplied by , divide by on both sides.

$\frac{I}{PR}=\frac{PRT}{PR}$

$\frac{I}{PR}=T$

Therefore, the equation solved for is,

$T=\frac{I}{PR}$

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Question

To calculate the volume of a rectangle the following formula is used.

Solve the equation for .

Answer

This question is asking to rearrange this formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since is being multiplied by , divide by on both sides.

$\frac{V}{wh}=\frac{lwh}{wh}$

$\frac{V}{wh}=l$

Therefore, the equation solved for is,

$l=\frac{V}{wh}$

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Question

To calculate the volume of a rectangle the following formula is used.

Solve the equation for .

Answer

This question is asking to rearrange this formula to highlight a quantity of interest, specifically .

To rearrange the formula, perform algebraic operations. Recall that whatever operation that is done to one side needs to also be done on the other side in order to keep the equation balanced.

Since is being multiplied by , divide by on both sides.

$\frac{V}{wl}=\frac{lwh}{wl}$

$\frac{V}{wl}=h$

Therefore, the equation solved for is,

$h=\frac{V}{wl}$

Compare your answer with the correct one above

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