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Algebra II : Equations

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Example Questions

Example Question #461 : Equations

Solve for x and y.

x-y=15

2x+2y=14

Possible Answers:

$x=11\text{ and }y=4$

$x=-11\text{ and }y=-4$

$x=5\text{ and }y=-6$

$x=11\text{ and }y=-4$

$x=5\text{ and }y=6$

Correct answer:

$x=11\text{ and }y=-4$

Explanation:

Solve for x and y by using substitution. Solve the first equation for x.

x=15+y

Next substitute x into the second equation and simplify,

2(15+y)+2y=14

30+2y+2y=14

4y=-16

y=-4

Now use the value for y to solve for x.

x=15+(-4)=11

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Example Question #462 : Equations

Solve for x and y.

3x-5y=37

2x+6y=-22

Possible Answers:

$x=2\text{ and }y=-5$

$x=4\text{ and }y=1$

$x=2\text{ and }y=-3$

$x=2\text{ and }y=3$

$x=4\text{ and }y=-5$

Correct answer:

$x=4\text{ and }y=-5$

Explanation:

Solve for x and y by using substitution. Solve the first equation for x.

$x=\frac{37+5y}{3}$

Next substitute x into the second equation and simplify,

$2(\frac{37+5y}{3})+6y=-22$

24.666+3.333y+6y=-22

9.333y=-46.666

y=-5

Now use the value for y to solve for x.

$x=\frac{37+5(-5)}{3}=4$

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Example Question #461 : Equations

Solve for x and y.

2x-3y=-13

5x+9y=83

Possible Answers:

$x=5\text{ and }y=2$

$x=-4\text{ and }y=-7$

$x=4\text{ and }y=-7$

$x=2\text{ and }y=5$

$x=4\text{ and }y=7$

Correct answer:

$x=4\text{ and }y=7$

Explanation:

Solve for x and y by using substitution. Solve the first equation for x.

$x=\frac{-13+3y}{2}$

Next substitute x into the second equation and simplify,

$5(\frac{-13+3y}{2})+9y=83$

-32.5+7.5y+9y=83

16.5y=115.5

y=7

Now use the value for y to solve for x.

$x=\frac{-13+3(7)}{2}=4$

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Example Question #331 : Solving Equations

Solve for x and y.

3x-5y=14

0.5x+0.5y=2.5

Possible Answers:

$x=10\text{ and }y=-5$

$x=-10\text{ and }y=-5$

$x=10\text{ and }y=5$

$x=-4\text{ and }y=2$

$x=4\text{ and }y=-2$

Correct answer:

$x=10\text{ and }y=-5$

Explanation:

Solve for x and y by using substitution. Solve the first equation for x.

$x=\frac{35+y}{3}$

Next substitute x into the second equation and simplify,

$0.5(\frac{35+y}{3})+0.5y=2.5$

5.833+0.166y+0.5y=2.5

0.666y=-3.333

y=-5

Now use the value for y to solve for x.

$x=\frac{35+(-5)}{3}=10$

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Example Question #811 : Basic Single Variable Algebra

Solve for x and y.

4x-5y=14

7x-12y=18

Possible Answers:

$x=-2\text{ and }y=4$

$x=11\text{ and }y=8$

$x=4\text{ and }y=6$

$x=5\text{ and }y=-1$

$x=6\text{ and }y=2$

Correct answer:

$x=6\text{ and }y=2$

Explanation:

Solve for x and y by using substitution. Solve the first equation for x.

$x=\frac{14+5y}{4}$

Next substitute x into the second equation and simplify,

$7(\frac{14+5y}{4})+-12y=18$

24.5+8.75y-12y=18

-3.25y=-6.5

y=2

Now use the value for y to solve for x.

$x=\frac{14+5(2)}{4}=6$

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Example Question #466 : Equations

Solve for given: 2x+8+10x=13x+10

Possible Answers:

1/2

-1/2

25/18

Correct answer:

Explanation:

In order to solve an equation for an unknown it must first be simplified by combining like terms.

In the case of 2x+8+10x=13x+10

Terms can be organized into those which have the variable in them and those that don't.

First 10x can be added to giving 12x , while is subtracted from , giving you:

12x=13x+2

Subtracting 13x from both sides gives you:

-x=2

In order to solve for you can multiply both sides of the equation by which gives you the final answer of .

This can be double checked by plugging it into the original equation:

2x+8+10x=13x+10

2*(-2)+8+10*(-2)=13*(-2)+10

-4+8-20=-26+10

-16=-16

thereby proving is a valid answer

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Example Question #467 : Equations

Solve for ,

$\sqrt{x+1}+3x = 2$

Possible Answers:

$x=\frac{7+ \sqrt{13}}{5}$

$x=\left \{\frac{7- \sqrt{13}}{5}\: \: \: ,\: \: \: \frac{7+ \sqrt{13}}{5} \right \}$

No solutions

$x=\left \{\frac{13- \sqrt{61}}{18}\: \: \: ,\: \: \: \frac{13+ \sqrt{61}}{18} \right \}$

$x=\frac{13- \sqrt{61}}{18}$

Correct answer:

$x=\frac{13- \sqrt{61}}{18}$

Explanation:

$\sqrt{x+1}+3x = 2$ (1)

When solving a radical equation the fist step is always to isolate the radical. Subtracting from both sides of equation (1).

$\sqrt{x+1} = 2-3x$

Square both sides and expand the right side,

x+1 = (2-3x)^2

x+1 =4-12x+9x^2

Collect all like-terms on onto one side of the equation and use the quadratic formula to find the roots:

9x^2 - 13x +3 = 0 (2)

____________________________________________________________

Reminder

Recall the general solution for the quadratic equation,

ax^2+bx+c=0

$x = \frac{-b\pm\sqrt{b^2-4ca}}{2a}$ (3)

____________________________________________________________

Use equation (3) to write the solutions to equation (2) and simplify:

$x = \frac{-(-13)\pm \sqrt{(-13)^2-4(3)(9)}}{2(9)}$

$x = \frac{13\pm \sqrt{61}}{18}$

Therefore, the roots to the quadratic equation (2) are:

$x=\left \{\frac{13- \sqrt{61}}{18}\: \: \: ,\: \: \: \frac{13+ \sqrt{61}}{18} \right \}$

These represent two possible solutions to equation (1). We must check both of them. This is because one of the steps in solving the original equation involved a squaring operation, which can produce fictitious solutions.

$\sqrt{\frac{13- \sqrt{61}}{18}+1}\: \: \: \: +\: \: \: 3\frac{13- \sqrt{61}}{18} \: \: \: =\: \: \: 2$

$\sqrt{\frac{13+ \sqrt{61}}{18}+1}\: \: \: \: +\: \: \: 3\frac{13+\sqrt{61}}{18} \: \: \: \neq \: \: \: 2$

Therefore, the only solution for the for equation (1) is:

$x=\frac{13- \sqrt{61}}{18}$

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Example Question #811 : Basic Single Variable Algebra

Solve 3=x-3+5x for .

Possible Answers:

x=0

x=3

x=1

x=2

x=-1

Correct answer:

x=1

Explanation:

Putting like variables on each side we get 6=6x , and divide each side by , getting x=1 .

$\\3=x-3+5x \\3=-3+6x \\3+3=-3+3+6x \\6=6x \\1=x$

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Example Question #2 : Systems Of Equations

Solve the system of equations:

4x-6y=12

2x+2y=6

Possible Answers:

x=1, y =2

y=0, x=3

y=-2, x=4

x=0, y=3

x=6,y=2

Correct answer:

y=0, x=3

Explanation:

Solve using elimination:

4x-6y =12

2(2x+2y=6) multiply the 2nd equation by two to make elimination possible
________________
4x-6y=12
4x+4y=12 subtract 2nd equation from the first to solve for
________________
-10y=0
${\color{Red} y=0}$

Substitute into either equation to solve for

2x+2(0)=6
2x=6
${\color{Red} x=3}$

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Example Question #461 : Equations

Nick’s sister Sarah is three times as old as him, and in two years will be twice as old as he is then. How old are they now?

Possible Answers:

Nick is 4, Sarah is 12

Nick is 2, Sarah is 6

Nick is 3, Sarah is 9

Nick is 4, Sarah is 8

Nick is 5, Sarah is 15

Correct answer:

Nick is 2, Sarah is 6

Explanation:

Step 1: Set up the equations

Let = Nick's age now
Let = Sarah's age now

The first part of the question says "Nick's sister is three times as old as him". This means:

s=3n

The second part of the equation says "in two years, she will be twice as old as he is then). This means:

s+2=2(n+2)

Add 2 to each of the variables because each of them will be two years older than they are now.

Step 2: Solve the system of equations using substitution

Substitute for in the second equation. Solve for

3n+2=2(n+2)

3n+2=4n+4

n+2=4

${\color{Red} n=2}$

Plug into the first equation to find

s=3(2)

${\color{Red} s=6}$

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Algebra II : Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #461 : Equations

Example Question #462 : Equations

Example Question #461 : Equations

Example Question #331 : Solving Equations

Example Question #811 : Basic Single Variable Algebra

Example Question #466 : Equations

Example Question #467 : Equations

Example Question #811 : Basic Single Variable Algebra

Example Question #2 : Systems Of Equations

Example Question #461 : Equations

All Algebra II Resources

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