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Common Core: High School - Algebra : High School: Algebra

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

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All Common Core: High School - Algebra Resources

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Example Questions

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,

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=r+y

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

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,

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

Report an Error

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,

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

Report an Error

Example Question #452 : High School: Algebra

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

V=lwh

Solve the equation for .

Possible Answers:

l=V-wh

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

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

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

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

Correct answer:

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

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.

V=lwh

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

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

Therefore, the equation solved for is,

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

Report an Error

Example Question #453 : High School: Algebra

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

V=lwh

Solve the equation for .

Possible Answers:

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

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

h=V-wl

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

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

Correct answer:

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

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.

V=lwh

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

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

Therefore, the equation solved for is,

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

Report an Error

Example Question #1 : Reasoning With Equations & Inequalities

Solve for .

x+2x+1=8+4x

Possible Answers:

x=-6

x=-9

x=9

x=-7

x=7

Correct answer:

x=-7

Explanation:

To solve for , first combine like terms.

x+2x+1=8+4x

On the left-hand side of the equation there are two terms that contain . Therefore, add and together.

3x+1=8+4x

Now, move the term from the left-hand side to the right-hand side. To accomplish this, subtract from both sides.

3x+1=8+4x

-3x -3x

_____________________

1=8+x

Next, to isolate , subtract the constant from the right-hand side of the equation to the left-hand side.

1=8+x

______________

-7=x

Report an Error

Example Question #2 : Reasoning With Equations & Inequalities

Solve for .

3x-2x+1=8+4x

Possible Answers:

x=-21

$x=\frac{3}{7}$

$x=\frac{7}{3}$

$x=-\frac{7}{3}$

$x=-\frac{3}{7}$

Correct answer:

$x=-\frac{7}{3}$

Explanation:

To solve for , first combine like terms.

3x-2x+1=8+4x

On the left-hand side of the equation there are two terms that contain . Therefore, subtract from .

x+1=8+4x

Now, move the term from the left-hand side to the right-hand side. To accomplish this, subtract from both sides.

x+1=8+4x

_____________________

1=8+3x

Next, subtract the constant from the right-hand side of the equation to the left-hand side.

1=8+3x

______________

-7=3x

Finally divide each side by three to solve for .

$\\\frac{-7}{3}=\frac{3x}{3} \\\\-\frac{7}{3}=x$

Report an Error

Example Question #1 : Reasoning With Equations & Inequalities

Solve for .

x+2x+3=8+4x

Possible Answers:

x=11

x=-11

x=-5

x=24

x=5

Correct answer:

x=-5

Explanation:

To solve for , first combine like terms.

x+2x+3=8+4x

On the left-hand side of the equation there are two terms that contain . Therefore, add and together.

3x+3=8+4x

Now, move the term from the left-hand side to the right-hand side. To accomplish this, subtract from both sides.

3x+3=8+4x

-3x -3x

_____________________

3=8+x

Next, to isolate , subtract the constant from the right-hand side of the equation to the left-hand side.

3=8+x

______________

-5=x

Report an Error

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : High School: Algebra

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

All Common Core: High School - Algebra Resources

Example Questions

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

Example Question #451 : High School: Algebra

Example Question #42 : Creating Equations✭

Example Question #43 : Creating Equations✭

Example Question #44 : Creating Equations✭

Example Question #452 : High School: Algebra

Example Question #453 : High School: Algebra

Example Question #1 : Reasoning With Equations & Inequalities

Example Question #2 : Reasoning With Equations & Inequalities

Example Question #1 : Reasoning With Equations & Inequalities

All Common Core: High School - Algebra Resources

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