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Algebra 1 : Systems of Equations

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Example Questions

Example Question #11 : Equations / Solution Sets

Solve for :

$\sqrt{x} + \sqrt{x -5} = 5$

Possible Answers:

x = -4

$x = -4 \textrm{ or }x = 9$

x=9

The equation has no solution.

$x = 4 \textrm{ or }x = 9$

Correct answer:

x=9

Explanation:

Move one radical to the other side, then square, thereby yielding an equation with only one radical.

$\sqrt{x} + \sqrt{x -5} = 5$

$\sqrt{x} + \sqrt{x -5} -\sqrt{x} = 5-\sqrt{x}$

$\sqrt{x -5} = 5-\sqrt{x}$

$\left (\sqrt{x -5} \right ) ^{2} =\left ( 5-\sqrt{x }\right ) ^{2}$

$x -5= 5 ^{2} - 2 \cdot 5 \cdot \sqrt{x } + \left ( \sqrt{x }\right ) ^{2}$

$x -5= 25 - 10 \sqrt{x } + x$

Isolate the radical on one side, then square.

$x -5-x -25= 25 - 10 \sqrt{x } + x -x -25$

$-30 = - 10 \sqrt{x }$

$-30 \div (-10) = - 10 \sqrt{x } \div (-10)$

$3= \sqrt{x }$

$3^{2} =\left ( \sqrt{x } \right )^{2}$

9 = x

Substitution confirms this to be the only solution.

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Example Question #1 : How To Find A Solution Set

If the area of a rectangle is 100 square feet and the width is 20 feet, then what is the perimeter?

Possible Answers:

80 feet

20 feet

50 feet

30 feet

Correct answer:

50 feet

Explanation:

The area of a rectangle is A=L*W , where A is the area, L is the length, and W is the width. The perimeter is given by P=2L+2W . We know that A=100 and W=20 . We can solve for L using $L=\frac{A}{W}=\frac{100}{20}=5.$ The perimeter is then 2*5+2*20=50 feet.

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Example Question #2 : How To Find A Solution Set

Solve for :

$\sqrt[3]{2n-8} = -4$

Possible Answers:

n = -36

n = 32

n = -28

$n = -28\textrm{ or }n = -36$

The equation has no solution.

Correct answer:

n = -28

Explanation:

Cube both sides of the equation to form a linear equation, then solve:

$\sqrt[3]{2n-8} = -4$

$\dpi{150} \left (\sqrt[3]{2n-8} \right ) ^{3}= \left (-4 \right )^{3}$

2n-8 = -64

2n-8 + 8 = -64 + 8

2n = -56

$2n \div 2 = -56\div 2$

n = -28

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Example Question #131 : Systems Of Equations

Solve this system of equations.

7x+4y+9z=0

8x-3y-6z=12

2x-y-3z=5

Possible Answers:

x=-2 , y=-2 , z=2

x=-1 , y=2 , $z=-\frac{5}{3}$

$x=\frac{1}{2}$ , $y=\frac{3}{2}$ , z=-3

x=1 , y=2 , $z=-\frac{5}{3}$

x=5 , y=-9 , z=3

Correct answer:

x=1 , y=2 , $z=-\frac{5}{3}$

Explanation:

Equation 1: 7x+4y+9z=0

Equation 2: 8x-3y-6z=12

Equation 3: 2x-y-3z=5

Adding the terms of the first and second equations together will yield 15x+y+3z=12 .

Then, add that to the third equation so that the y and z terms are eliminated. You will get 17x=17 .

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

7+4y+9z=0

8-3y-6z=12

2-y-3z=5

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

2-y-3z=5

-y=3z+5-2

y=-3z-3

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

7+4y+9z=0

7+4(-3z-3) +9z=0

7-12z-12+9z=0

-5-3z=0

$z=-\frac{5}{3}$

We can use this z value to find y

y=-3z-3

$y=-3\left ( -\frac{5}{3}\right )-3$

y=2

So the solution set is x = 1, y = 2, and z = –5/3.

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Example Question #2 : Solving Equations

Solve for : $x^{2}-144=0$

Possible Answers:

x=8,-4

x=12,-12

x=14,-14

x=10,-14

Correct answer:

x=12,-12

Explanation:

To solve this problem we can first add 144 to each side of the equation yielding $x^{2}=144$

Then we take the square root of both sides to get $\sqrt{x^{2}}=\sqrt{144}$

Then we calculate the square root of 144 which is $x=\pm12$ .

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Example Question #1 : Graphing Linear Functions

Determine where the graphs of the following equations will intersect.

3x+2y=-1

-x-3y=5

Possible Answers:

(2,-3)

(-2,1)

(1,-2)

(-8,1)

Correct answer:

(1,-2)

Explanation:

We can solve the system of equations using the substitution method.

3x+2y=-1

-x-3y=5

Solve for in the second equation.

-x-3y=5

-3y=5+x

-3y-5=x

Substitute this value of into the first equation.

3(-3y-5)+2y=-1

Now we can solve for .

-9y-15+2y=-1

-7y-15=-1

-7y=14

$y=\frac{14}{-7}=-2$

Solve for using the first equation with this new value of .

3x+2(-2)=-1

3x-4=-1

3x=3

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

The solution is the ordered pair (1,-2) .

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Example Question #132 : Systems Of Equations

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Possible Answers:

$\left \{ 2,3\right \}$

$\left \{ -3, -2\right \}$

$\left \{ -3,3\right \}$

$\left \{ -2,2\right \}$

$\left \{ -3, -2, 2, 3\right \}$

Correct answer:

$\left \{ -2,2\right \}$

Explanation:

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as (u+?)(u+?) , replacing the two question marks with integers with product and sum 5; these integers are 9,-4 .

(u+9)(u-4) = 0

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

$(x^{2}+9)(x+2)(x-2) = 0$

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

$x+2 = 0 \Rightarrow x = - 2$

The solution set is $\left \{ -2, 2\right \}$ .

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Example Question #133 : Systems Of Equations

Solve for :

-3x + 12 = -3

Possible Answers:

Correct answer:

Explanation:

Subtract 12 from both sides:

–3x = –15

Divide both sides by –3:

x = 5

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Example Question #134 : Systems Of Equations

Solve for :

x(x - 2) = x^2 -8

Possible Answers:

Correct answer:

Explanation:

Distribute the x through the parentheses:

x² –2x = x² – 8

Subtract x² from both sides:

–2x = –8

Divide both sides by –2:

x = 4

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Example Question #135 : Systems Of Equations

Solve for :

$\dpi{130} \sqrt[3]{100-x} - 5 = 0$

Possible Answers:

The equation has no solution.

$x = -25 \textrm{ or } x = 225$

x = 225

x = 125

x = -25

Correct answer:

x = -25

Explanation:

Add 5 to both sides to isolate the cube root:

$\dpi{130} \sqrt[3]{100-x} - 5+ 5 = 0+ 5$

$\dpi{130} \sqrt[3]{100-x} = 5$

Cube both sides:

$\dpi{130} \left (\sqrt[3]{100-x} \right) ^{3}= 5 ^{3}$

100-x = 125

To isolate , move it to the right side of the equation. We choose the right side instead of the left side so as to make positive:

100-x +x= 125+x

100 = 125+x

Subtract 125 from both sides to isolate :

100 - 125 = 125 - 125 +x

-25 = x

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Algebra 1 : Systems of Equations

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #11 : Equations / Solution Sets

Example Question #1 : How To Find A Solution Set

Example Question #2 : How To Find A Solution Set

Example Question #131 : Systems Of Equations

Example Question #2 : Solving Equations

Example Question #1 : Graphing Linear Functions

Example Question #132 : Systems Of Equations

Example Question #133 : Systems Of Equations

Example Question #134 : Systems Of Equations

Example Question #135 : Systems Of Equations

All Algebra 1 Resources

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