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

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Example Questions

Example Question #1 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

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

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

Solve this equation for y_2 $\displaystyle y_2$.

Possible Answers:

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

$\displaystyle y_2=x_2m-x_1+y_1$

$\displaystyle y_2=(x_2-x_1)+y_1m$

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

$\displaystyle y_2=x_2-x_1m+y_1$

Correct answer:

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

Explanation:

This question is asking to rearrange the slope formula to highlight a quantity of interest, specifically y_2 $\displaystyle y_2$.

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 $\displaystyle (x_2-x_1)$ 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.

$\$x_2-x_1)\cdot m=\frac{y_2-y_1}{x_2-x_1}\cdot (x_2-x_1) \\\\(x_2-x_1)m=y_2-y_1$ \(\displaystyle \\(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 y_1 $\displaystyle y_1$ to both sides of the equation.

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

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

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Example Question #2 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

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

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

Solve this equation for y_1 $\displaystyle y_1$.

Possible Answers:

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

$\displaystyle y_1=-(x_2-x_1)m+y_2m$

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

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

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

Correct answer:

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

Explanation:

This question is asking to rearrange the slope formula to highlight a quantity of interest, specifically y_1 $\displaystyle y_1$.

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.

$\$x_2-x_1)\cdot m=\frac{y_2-y_1}{x_2-x_1}\cdot (x_2-x_1) \\\\(x_2-x_1)m=y_2-y_1$ \(\displaystyle \\(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 y_2 $\displaystyle y_2$ to both sides of the equation.

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

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

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

Therefore, the equation solved for y_1 $\displaystyle y_1$ is,

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

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Example Question #3 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

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

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

Solve this equation for x_2 $\displaystyle x_2$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

This question is asking to rearrange the slope formula to highlight a quantity of interest, specifically x_2 $\displaystyle x_2$.

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.

$\$x_2-x_1)\cdot m=\frac{y_2-y_1}{x_2-x_1}\cdot (x_2-x_1) \\\\(x_2-x_1)m=y_2-y_1$ \(\displaystyle \\(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 $\displaystyle m$ on both sides.

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

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

Now add x_1 $\displaystyle x_1$ to both sides.

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

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

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Example Question #4 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

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

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

Solve this equation for x_1 $\displaystyle x_1$.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

This question is asking to rearrange the slope formula to highlight a quantity of interest, specifically x_1 $\displaystyle x_1$.

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.

$\$x_2-x_1)\cdot m=\frac{y_2-y_1}{x_2-x_1}\cdot (x_2-x_1) \\\\(x_2-x_1)m=y_2-y_1$ \(\displaystyle \\(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 $\displaystyle m$ on both sides.

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

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

Now subtract x_2 $\displaystyle x_2$ from both sides.

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

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

Finally, divide by negative one.

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

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Example Question #3 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

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

$\displaystyle x^2+y^2=r^2$

Solve the equation for $\displaystyle r$.

Possible Answers:

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

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

r=x+y $\displaystyle r=x+y$

$\displaystyle r=x^2+y^2$

r=x-y $\displaystyle r=x-y$

Correct answer:

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

Explanation:

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

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 $\displaystyle r$ take the square root of each side.

$\displaystyle x^2+y^2=r^2$

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

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

Therefore, the equation solved for $\displaystyle r$ is

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

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Example Question #4 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

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

$\displaystyle x^2+y^2=r^2$

Solve the equation for $\displaystyle x$.

Possible Answers:

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

x=r-y $\displaystyle x=r-y$

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

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

x=r+y $\displaystyle x=r+y$

Correct answer:

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

Explanation:

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

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 $\displaystyle x$ subtract y^2 $\displaystyle y^2$ from both sides and then take the square root.

$\displaystyle x^2+y^2=r^2$

$\displaystyle x^2+y^2-y^2=r^2-y^2$

$\displaystyle x^2=r^2-y^2$

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

Therefore, the equation solved for $\displaystyle x$ is

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

Report an Error

Example Question #1 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

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

$\displaystyle x^2+y^2=r^2$

Solve the equation for $\displaystyle y$.

Possible Answers:

y=r+x $\displaystyle y=r+x$

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

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

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

y=r-x $\displaystyle y=r-x$

Correct answer:

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

Explanation:

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

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 $\displaystyle y$ subtract x^2 $\displaystyle x^2$ from both sides and then take the square root.

$\displaystyle x^2+y^2=r^2$

$\displaystyle x^2+y^2-x^2=r^2-x^2$

$\displaystyle y^2=r^2-x^2$

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

Therefore, the equation solved for $\displaystyle y$ is

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

Report an Error

Example Question #2 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

To calculate simple interest the following formula is used.

I=PRT $\displaystyle I=PRT$

Solve the equation for $\displaystyle P$.

Possible Answers:

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

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

P=I-RT $\displaystyle P=I-RT$

P=IRT $\displaystyle P=IRT$

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

Correct answer:

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

Explanation:

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

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 $\displaystyle P$ is being multiplied by $\displaystyle RT$, divide by $\displaystyle RT$ on both sides.

I=PRT $\displaystyle I=PRT$

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

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

Therefore, the equation solved for $\displaystyle P$ is,

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

Report an Error

Example Question #9 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

To calculate simple interest the following formula is used.

I=PRT $\displaystyle I=PRT$

Solve the equation for $\displaystyle R$.

Possible Answers:

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

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

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

R=IPT $\displaystyle R=IPT$

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

Correct answer:

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

Explanation:

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

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 $\displaystyle R$ is being multiplied by $\displaystyle PT$, divide by $\displaystyle PT$ on both sides.

I=PRT $\displaystyle I=PRT$

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

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

Therefore, the equation solved for $\displaystyle R$ is,

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

Report an Error

Example Question #10 : Rearrange Formulas And Solve Equations: Ccss.Math.Content.Hsa Ced.A.4

To calculate simple interest the following formula is used.

I=PRT $\displaystyle I=PRT$

Solve the equation for $\displaystyle T$.

Possible Answers:

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

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

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

T=I-PR $\displaystyle T=I-PR$

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

Correct answer:

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

Explanation:

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

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 $\displaystyle T$ is being multiplied by $\displaystyle PR$, divide by $\displaystyle PR$ on both sides.

I=PRT $\displaystyle I=PRT$

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

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

Therefore, the equation solved for $\displaystyle T$ is,

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

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

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