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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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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #31 : Creating Equations✭

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 y_2 .

Possible Answers:

y_2=(x_2-x_1)m

y_2=x_2-x_1m+y_1

y_2=(x_2-x_1)m+y_1

y_2=(x_2-x_1)+y_1m

y_2=x_2m-x_1+y_1

Correct answer:

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 .

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 (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.

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

Next, add y_1 to both sides of the equation.

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

Therefore, the equation solved for y_2 is

y_2=(x_2-x_1)m+y_1

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Example Question #32 : Creating Equations✭

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 y_1 .

Possible Answers:

y_1=-(x_2-x_1)m+y_2m

y_1=-(x_2-x_1)m+y_2

y_1=-(x_2-x_1)m-y_2

y_1=(x_2-x_1)m+y_2

y_1=(x_2-x_1)m-y_2

Correct answer:

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 .

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 (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.

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

Next, subtract y_2 to both sides of the equation.

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

Now, divide by negative one on both sides.

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

-(x_2-x_1)m+y_2=y_1

Therefore, the equation solved for y_1 is,

y_1=-(x_2-x_1)m+y_2

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Example Question #33 : Creating Equations✭

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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 (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.

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

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 x_1 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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Example Question #34 : Creating Equations✭

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 .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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 (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.

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

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 x_2 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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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,

x^2+y^2=r^2

Solve the equation for .

Possible Answers:

r=x-y

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

r=x+y

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

r=x^2+y^2

Correct answer:

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

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 take the square root of each side.

x^2+y^2=r^2

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

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

Therefore, the equation solved for is

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

Report an Error

Example Question #42 : Creating Equations✭

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

x^2+y^2=r^2

Solve the equation for .

Possible Answers:

x=r-y

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

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

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

x=r+y

Correct answer:

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

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 y^2 from both sides and then take the square root.

x^2+y^2=r^2

x^2+y^2-y^2=r^2-y^2

x^2=r^2-y^2

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

Therefore, the equation solved for is

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

Report an Error

Example Question #451 : High School: Algebra

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

x^2+y^2=r^2

Solve the equation for .

Possible Answers:

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

y=r-x

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

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

y=r+x

Correct answer:

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

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 x^2 from both sides and then take the square root.

x^2+y^2=r^2

x^2+y^2-x^2=r^2-x^2

y^2=r^2-x^2

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

Therefore, the equation solved for is

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

Report an Error

Example Question #42 : Creating Equations✭

To calculate simple interest the following formula is used.

I=PRT

Solve the equation for .

Possible Answers:

P=I-RT

P=IRT

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

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

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

Correct answer:

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

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.

I=PRT

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

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

Therefore, the equation solved for is,

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

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Example Question #43 : Creating Equations✭

To calculate simple interest the following formula is used.

I=PRT

Solve the equation for .

Possible Answers:

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

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

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

R=IPT

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

Correct answer:

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

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.

I=PRT

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

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

Therefore, the equation solved for is,

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

Report an Error

Example Question #44 : Creating Equations✭

To calculate simple interest the following formula is used.

I=PRT

Solve the equation for .

Possible Answers:

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

T=I-PR

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

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

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

Correct answer:

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

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.

I=PRT

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

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

Therefore, the equation solved for is,

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

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

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #31 : Creating Equations✭

Example Question #32 : Creating Equations✭

Example Question #33 : Creating Equations✭

Example Question #34 : Creating Equations✭

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

Example Question #42 : Creating Equations✭

Example Question #451 : High School: Algebra

Example Question #42 : Creating Equations✭

Example Question #43 : Creating Equations✭

Example Question #44 : Creating Equations✭

All Common Core: High School - Algebra Resources

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