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

To calculate simple interest the following formula is used.

Solve the equation for .

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

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

$T=\frac{I}{P}-R$

$T=\frac{I}{R}-P$

Explanation

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,

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

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

Solve the equation for .

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

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

$l=\frac{V}{w}-h$

$l=\frac{V}{h}-w$

Explanation

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,

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

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

Solve the equation for .

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

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

$h=\frac{V}{w}-l$

$h=\frac{V}{l}-w$

Explanation

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,

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

To calculate simple interest the following formula is used.

Solve the equation for .

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

$R=\frac{I}{P}-T$

$R=\frac{I}{T}-P$

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

Explanation

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,

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

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 .

Explanation

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

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 .

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

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

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

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

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

Explanation

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.

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

To calculate simple interest the following formula is used.

Solve the equation for .

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

$P=\frac{I}{R}-T$

$P=\frac{I}{T}-R$

Explanation

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,

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

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

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

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

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

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

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

Explanation

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.

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

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

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

Solve the equation for .

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

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

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

Explanation

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

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

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

Solve the equation for .

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

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

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

Explanation

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.