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Algebra 1 : How to find the solution for a system of equations

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Example Questions

Example Question #142 : Systems Of Equations

A cube has a volume of 216 $cm^{3}$ . If its width is $3x^{2}$ , its length is -0.5x , and its height is $x^{3}$ +14 , find .

Possible Answers:

Correct answer:

Explanation:

Since the object in question is a cube, each of its sides must be the same length. Therefore, to get a volume of 216 $cm^{3}$ , each side must be equal to the cube root of 216 , which is cm.

We can then set each expression equal to .

The first expression (3x^2^ -6) can be solved by either or , but the other two expressions make it evident that the solution is x = -2 .

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Example Question #151 : Systems Of Equations

3x + y = 8

2x + 4y = 12

Solve the system for and .

Possible Answers:

x = 2, y = 2

x = 10, y = -2

x = 2, y = 0

x = 3, y = -1

x = -4, y = 20

Correct answer:

x = 2, y = 2

Explanation:

The most simple method for solving systems of equations is to transform one of the equations so it allows for the canceling out of a variable. In this case, we can multiply 3x + y = 8 by (-4) to get -12x - 4y = -32 .

Then, we can add 2x + 4y = 12 to this equation to yield -10x = -20 , so x = 2 .

We can plug that value into either of the original equations; for example, 3(2) )+ y = 8 .

So, y = 2 as well.

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Example Question #2 : How To Find The Solution For A System Of Equations

What is the solution to the following system of equations:

3x+5y = 15

3x+3y = 15

Possible Answers:

$x=\frac{1}{3},y=3$

$x=3,y=\frac{6}{5}$

x=3,y=2

$x=2,y=\frac{13}{4}$

x = 5, y = 0

Correct answer:

x = 5, y = 0

Explanation:

By solving one equation for , and replacing in the other equation with that expression, you generate an equation of only 1 variable which can be readily solved.

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Example Question #152 : Systems Of Equations

Solve this system of equations for :

3x + 5y = 1

6x-y = 24

Possible Answers:

$3\frac{2}{3}$

None of the other choices are correct.

$2\frac{2}{3}$

$3\frac{1}{3}$

Correct answer:

$3\frac{2}{3}$

Explanation:

Multiply the bottom equation by 5, then add to the top equation:

6x-y = 24

$5 \left (6x-y \right ) =5\cdot 24$

30x-5y =120

$\underline{\textrm{\; } 3x + 5y = \; \; 1}$

$33x \;\;\;\;\;\; \; =121$

$x = \frac{121}{33} = \frac{11}{3} = 3 \frac{2}{3}$

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Example Question #153 : Systems Of Equations

Solve this system of equations for :

3x + 5y = 23

6x-y = -9

Possible Answers:

$y = 2 \frac{1}{2}$

y = 3

None of the other choices are correct.

$y = 3 \frac{1}{2}$

y = 5

Correct answer:

y = 5

Explanation:

Multiply the top equation by :

3x + 5y = 23

$-2 \cdot \left (3x + 5y \right )= -2 \cdot 23$

$-6x -10y \right )= -46$

Now add:

$6x-\; \; y = -9$

$\underline{-6x -10y = -46}$

-11y = - 55

$-11y \div (-11) = -55\div (-11)$

y = 5

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Example Question #163 : Expressions & Equations

Solve this system of equations for :

3x + 5y = 1

6x-y = 24

Possible Answers:

None of the other choices are correct.

$y = -1 \frac{3}{5}$

$y = -2 \frac{1}{5}$

$y = -1 \frac{4}{5}$

y = -2

Correct answer:

y = -2

Explanation:

Multiply the top equation by :

3x + 5y = 1

$-2 \cdot \left (3x + 5y \right )= -2 \cdot 1$

$-6x -10y \right )= -2$

Now add:

$\; \; 6x-\; \; \; y = 24$

$\underline{-6x -10y= -2} \right )$

-11y = 22

$-11y \div (-11) = 22\div (-11)$

y = -2

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Example Question #1 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Find the solution to the following system of equations.

a + 3b = 5

a-2b = 0

Possible Answers:

$a = 1,\ b = 1$

$a = 2,\ b = 2$

$a = 1,\ b = 2$

$a = 2,\ b = 3$

$a = 2,\ b = 1$

Correct answer:

$a = 2,\ b = 1$

Explanation:

a + 3b = 5

a-2b = 0

To solve this system of equations, use substitution. First, convert the second equation to isolate $\small a$ .

$a-2b = 0\rightarrow a=2b$

Then, substitute $\small 2b$ into the first equation for $\small a$ .

a + 3b = 5

(2b) + 3b = 5

Combine terms and solve for $\small b$ .

5b=5

b=1

Now that we know the value of $\small b$ , we can solve for $\small a$ using our previous substitution equation.

a=2b=2(1)=2

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Example Question #165 : Expressions & Equations

Find a solution for the following system of equations:

$\left\{\begin{matrix} x-2y-1=3\\ -x+2y+2=5 \end{matrix}\right.$

Possible Answers:

$\left ( 4,0 \right )$

infinitely many solutions

$\left ( 8,2 \right )$

no solution

Correct answer:

no solution

Explanation:

When we add the two equations, the and variables cancel leaving us with:

1=8 which means there is no solution for this system.

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Example Question #1 : How To Find The Solution For A System Of Equations

Solve for :

4x+3=9x+12-2x

Possible Answers:

None of the other answers

Correct answer:

Explanation:

First, combine like terms to get 4x+3=7x+12 . Then, subtract 12 and from both sides to separate the integers from the 's to get -9=3x . Finally, divide both sides by 3 to get x=-3 .

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Example Question #1 : How To Find The Solution For A System Of Equations

We have two linear functions:

y=-4x+8

y=3x-5

Find the coordinate at which they intersect.

Possible Answers:

none of these

$(x,y)=\left(\frac{5}{2},\frac{1}{7} \right )$

$(x,y)=\left(14,7 )$

$(x,y)=\left(-\frac{13}{7},-\frac{1}{9} \right )$

$(x,y)=\left(\frac{13}{7},\frac{4}{7} \right )$

Correct answer:

$(x,y)=\left(\frac{13}{7},\frac{4}{7} \right )$

Explanation:

We are given the following system of equations:

y=-4x+8

y=3x-5

We are to find and . We can solve this through the substitution method. First, substitute the second equation into the first equation to get

3x-5=-4x+8

Solve for by adding 4x to both sides

3x+4x-5=-4x+4x+8

7x-5=8

Add 5 to both sides

7x-5+5=8+5

7x=13

Divide by 7

$\frac{7x}{7}=\frac{13}{7}$

$x=\frac{13}{7}$

So $x=\frac{13}{7}$ . Use this value to find using one of the equations from our given system of equations. I think I'll use the first equation (can also use the second equation).

y=-4x+8

$y=-4\left (\frac{13}{7} \right )+8$

$y=-\frac{52}{7}+8$

$y=-\frac{52}{7}+\frac{56}{7}$

$y=\frac{4}{7}$

So the two linear functions intersect at

$(x,y)=\left(\frac{13}{7},\frac{4}{7} \right )$

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Algebra 1 : How to find the solution for a system of equations

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #142 : Systems Of Equations

Example Question #151 : Systems Of Equations

Example Question #2 : How To Find The Solution For A System Of Equations

Example Question #152 : Systems Of Equations

Example Question #153 : Systems Of Equations

Example Question #163 : Expressions & Equations

Example Question #1 : Solve Systems Of Two Linear Equations: Ccss.Math.Content.8.Ee.C.8b

Example Question #165 : Expressions & Equations

Example Question #1 : How To Find The Solution For A System Of Equations

Example Question #1 : How To Find The Solution For A System Of Equations

All Algebra 1 Resources

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