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College Algebra : Solving Equations and Inequallities

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Example Questions

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Example Question #9 : Linear Equations

Solve for .

7(x-2)+10=8(x+1)

Possible Answers:

x=4

x=-12

x=12

x=1

x=2

Correct answer:

x=-12

Explanation:

First distribute out each side of the equation.

7(x-2)+10 simplifies to 7x-4 .

Now for the right hand side,

8(x+1) becomes 8x+8 .

Now we equate both sides.

7x-4 =8x+8 ,

x=-12

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Example Question #10 : Linear Equations

Solve for .

3(x+2-2x)+5x=x+6

Possible Answers:

x=-12

x=0

x=2

x=4

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

Correct answer:

x=0

Explanation:

First, we need to simplify what's inside the parentheses.

x+2-2x=2-x

Now we continue to evaluate the left hand side.

3(2-x)+5x=2x+6

The right hand side does not need any reduction.

We set the two sides equal to each other.

2x+6=x+6

x=0

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Example Question #11 : Linear Equations

Solve the equation: $3x = -\frac{1}{18}$

Possible Answers:

$-\frac{1}{9}$

$-\frac{1}{54}$

$-\frac{1}{6}$

Correct answer:

$-\frac{1}{54}$

Explanation:

In order to isolate the x-variable, we will need to multiply both sides by one third.

$3x \cdot \frac{1}{3}= -\frac{1}{18} \cdot \frac{1}{3}$

Simplify both sides.

$x= -\frac{1}{54}$

The answer is: $-\frac{1}{54}$

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Example Question #11 : Solving Equations And Inequallities

Evaluate: -9x+6=7x-1

Possible Answers:

$-\frac{7}{16}$

$-\frac{7}{2}$

$-\frac{5}{16}$

$\frac{7}{16}$

Correct answer:

$\frac{7}{16}$

Explanation:

Add on both sides.

-9x+6+9x=7x-1+9x

6= -1+16x

Add one on both sides.

6+1= -1+16x+1

7=16x

Divide by 16 on both sides.

$\frac{7}{16}=\frac{16x}{16}$

The answer is: $\frac{7}{16}$

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Example Question #11 : Solving Equations And Inequallities

Express the following linear inequality in interval notation.

$7-2x\geq1$

Possible Answers:

$(-\infty ,-3)$

$[3,\infty )$

$(-\infty, -3]$

$(-\infty,3]$

$(-\infty,3)$

Correct answer:

$(-\infty,3]$

Explanation:

Upon solving for x, we find that x is less than or equal to 3. The left-hand term of the interval is negative infinity since any number less than 3 is in our set, and infinity always has a parenthesis around it. The right-hand term of the interval is 3 since it is the upper bound of our set. There is a bracket around it because 3 is included in our set (3 is less than or equal to 3). Remember when dividing or multiplying by a negative number in an inequality to reverse the direction of the inequality.

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Example Question #11 : Solving Equations And Inequallities

Solve: 2x+6=12

Possible Answers:

x=77

x=2

x=3

x=-4

Correct answer:

x=3

Explanation:

Step 1: Subtract 6 from both sides...

2x+6-6=12-6

Step 2: Divide by 2.

$\frac {2x}{2}=\frac {6}{2}$

Simplify:

x=3

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Example Question #11 : Solving Equations And Inequallities

Solve: 2x^2+16=24 (positive roots only)

Possible Answers:

Correct answer:

Explanation:

Step 1: Subtract from both sides

2x^2+16-16=24-16

Step 2: Simplify:

2x^2=8

Step 3: Divide..

$\frac {2x^2}{2}=\frac {8}{2}$

Step 4: Take the square root of both sides...

$\sqrt{x^2}=\sqrt{4}$

Step 5: Simplify and get the answer...

x=2

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Example Question #16 : Linear Equations

Solve: 4x+3=11

Possible Answers:

x=-3

x=1

x=2

x=-2

x=-3

Correct answer:

x=2

Explanation:

Move to the other side by subtracting it from both sides.

4x+3-3=11-3

Simplify:

4x=8

Divide by the coefficient, the number in front of x.

$\frac {4x}{4}=\frac {8}{4}$

Reduce:

x=2

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Example Question #1 : Equations With More Than One Variable

Solve for X:

Y=8X+3

Possible Answers:

$X=\frac{y+3}{8}$

$X=\frac{Y-8}{3}$

$X=\frac{8}{Y-3}$

$X=\frac{8}{Y+3}$

$X=\frac{y-3}{8}$

Correct answer:

$X=\frac{y-3}{8}$

Explanation:

In Y=8X+3 , if we're solving for x, we first need to get the "x" term isolated. We do this by subtracting 3 from both sides so:

Y=8X+3

becomes

Y-3=8X

Now we divide both sides by 8

$\frac{Y-3}{8}=X$

Re-written the answer becomes

$X=\frac{Y-3}{8}$

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Example Question #1 : Equations With More Than One Variable

Larry has a handful of dimes and quarters. In total, he has 14 coins with a value of $2.60. How many of each coin does he have?

Possible Answers:

6 Dimes

8 Quarters

7 Dimes

7 Quarters

10 Dimes

4 Quarters

8 Dimes

6 Quarters

9 Dimes

5 Quarters

Correct answer:

6 Dimes

8 Quarters

Explanation:

Since this problems has 2 variables (D-dimes and Q-quarters) we need 2 equations. Because Larry has 14 coins, the first equation can be written as:

D+Q=14

The value of those coins equals $2.60 or 260 cents. If Dimes are worth 10C and quarters are 25C, the next equation can be written as

10D+25Q=260

To solve this write both equations on top of each other

D+Q=14

10D+25Q=260

Now we eliminate 1 variable by multiplying 1 equation by the lowest common denominator (as a negative) and adding the equations together.

-10(D+Q=14) becomes

-10D-10Q=-140

adding the equations

-10D-10Q=-140

10D+25Q=260

-----------------------------------

15Q=120

now we solve for Q.

$Q=\frac{120}{15}=8$

Since we know Q, now we plug it back in to an equation and find D

D+8=14 --> D=6

Larry has 6 dimes and 8 quarters

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College Algebra : Solving Equations and Inequallities

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #9 : Linear Equations

Example Question #10 : Linear Equations

Example Question #11 : Linear Equations

Example Question #11 : Solving Equations And Inequallities

Example Question #11 : Solving Equations And Inequallities

Example Question #11 : Solving Equations And Inequallities

Example Question #11 : Solving Equations And Inequallities

Example Question #16 : Linear Equations

Example Question #1 : Equations With More Than One Variable

Example Question #1 : Equations With More Than One Variable

All College Algebra Resources

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