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Algebra 1 : Linear / Rational / Variable Equations

Study concepts, example questions & explanations for Algebra 1

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Example Questions

Example Question #31 : Linear / Rational / Variable Equations

Solve for x: 12x-5(x-2)=45

Possible Answers:

x=8

x=7

$x=\frac{55}{7}$

$x=\frac{7}{55}$

x=5

Correct answer:

x=5

Explanation:

12x-5(x-2)=45

Simplify the parenthesis:

12x-5x+10=45

Combine the terms with x's:

7x+10=45

Combine constants:

$7x=35\rightarrow \mathbf{x=5}$

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Example Question #152 : Equations / Inequalities

Solve for x: $2x=\frac{3}{2}-5$

Possible Answers:

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

$x=-\frac{7}{4}$

None of these

$x=\frac{4}{3}$

$x=-\frac{4}{7}$

Correct answer:

$x=-\frac{7}{4}$

Explanation:

$2x=\frac{3}{2}-5$

Simplify the right side:

$2x=\frac{3}{2}-\frac{10}{2}$

$2x=-\frac{7}{2}$

Isolate x:

$\frac{1}{2}*2x=-\frac{7}{2}*\frac{1}{2}$

$\mathbf{x=-\frac{7}{4}}$

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Example Question #32 : Linear / Rational / Variable Equations

Solve for

1=5+x

Possible Answers:

None of the other answers.

Correct answer:

Explanation:

1 = 5 + x

1-5 = 5+x-5

-4=x

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Example Question #1 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for :

6x-9-x=2x-6

Possible Answers:

Correct answer:

Explanation:

First. combine like terms to get

5x-9=2x-6 .

Then, add and subtract from both sides to separate the terms.

This gives you 3x=3 .

Finally, divide both sides by to get a solution of x=1 .

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Example Question #2 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for :

8(x+5)=16

Possible Answers:

$\frac{11}{8}$

None of the other answers

$-\frac{11}{8}$

Correct answer:

Explanation:

First, you must multiply the left side of the equation using the distributive property.

This gives you 8x+40=16 .

Next, subtract from both sides to get 8x=-24 .

Then, divide both sides by to get x=-3 .

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Example Question #33 : Linear / Rational / Variable Equations

$A\ \blacksquare\ B= \frac{A + B}{2}-\frac{2}{A-B}$

If $8\ \blacksquare\ B = 6$ , find the value of .

Possible Answers:

Correct answer:

Explanation:

The simplest way to solve this is to test the answer choices provided by plugging them in the equation. Start with the number in the middle, which in this case is 6. Replace every "B" in the equation with "6", and solve.

$\frac{8+(6)}{2}-\frac{2}{8-(6)}=\frac{14}{2}-\frac{2}{2}=7-1=6$

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Example Question #1 : How To Find The Solution To A Rational Equation With Lcd

Simplify the following:

$\frac{2x}{x+3} + \frac{4}{x}$

Possible Answers:

$\frac{2x^2+4x+12}{x(x+3)}$

$\frac{2x+4}{x(x+3)}$

$\frac{8x}{x^2+3x}$

$\frac{2x+4}{2x+3}$

Correct answer:

$\frac{2x^2+4x+12}{x(x+3)}$

Explanation:

Before you can add the two numerators from the fractions, the fractions must have a common denominator. The common denominator will be the Lowest Common Denominator (LCD). When you are dealing with variables, you get this by multiplying the two denominators together:

$\frac{2x}{x+3} + \frac{4}{x} = \frac{?}{x(x+3)}$

To follow through with this LCD, you must multiply each fraction by the other fraction's denominator so that they end up with common denominators. The first fraction needs to be multiplied by $\frac{x}{x}$ and the second fraction needs to be multiplied by $\frac{x+3}{x+3}$ . Note that both of these are equivalent to ONE, so the value of the fraction does not change:

$\frac{(2x)(x)}{(x+3)(x)} + \frac{(4)(x+3)}{(x)(x+3)} = \frac{2x^2}{x(x+3)} + \frac{4x+12}{x(x+3)}$

Now that the two fractions have common denominators, you can add the two numerators:

$\frac{2x^2+4x+12}{x(x+3)}$

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Example Question #2 : How To Find The Solution To A Rational Equation With Lcd

Simplify the following:

$\frac{4x}{7} - \frac{2x}{2x+4}$

Possible Answers:

$\frac{2x}{14x+28}$

$\frac{4x^2+15x}{7x+14}$

$\frac{4x^2+4}{14x+28}$

$\frac{4x^2+x}{7x+14}$

Correct answer:

$\frac{4x^2+x}{7x+14}$

Explanation:

Before you can subtract the two numerators from the fractions, the fractions must have a common denominator. The common denominator will be the Lowest Common Denominator (LCD). When you are dealing with variables, you get this by multiplying the two denominators together:

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

To follow through with this LCD, you must multiply each fraction by the other fraction's denominator so that they end up with common denominators. The first fraction needs to be multiplied by $\frac{2x+4}{2x+4}$ and the second fraction needs to be multiplied by $\frac{7}{7}$ . Note that both of these are equivalent to ONE, so the value of the fraction does not change:

$\frac{4x(2x+4)}{7(2x+4)} - \frac{2x(7)}{2x+4(7)} = \frac{8x^2+16x}{7(2x+4)} - \frac{14x}{7(2x+4)}$

Now that the two fractions have common denominators, you can subtract the two numerators. You can also distribute the denominator so you can simplify later:

$\frac{8x^2+2x}{14x+28}$

Finally you can divide out each term in the numerator and denominator by 2 to fully simplify the answer:

$\frac{4x^2+x}{7x+14}$

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Example Question #1 : Solving Rational Expressions

Simplify:

$\frac{4x}{x+3}+\frac{2x}{x+3}$

Possible Answers:

$\frac{6}{x}$

$\frac{8x}{x+3}$

$\frac{6x}{x+3}$

Correct answer:

$\frac{6x}{x+3}$

Explanation:

$\frac{4x}{x+3}+\frac{2x}{x+3}$

Because the two rational expressions have the same denominator, we can simply add straight across the top. The denominator stays the same.

Therefore the answer is $\frac{6x}{x+3}$ .

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Example Question #2 : Solving Rational Expressions

Solve for :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

Possible Answers:

$x = 1 \frac{2}{5} \textrm{ or } x = 4$

$x = 4\textrm{ or }x = 7$

x = 4

x = 7

$x = 1 \frac{2}{5}$

Correct answer:

$x = 1 \frac{2}{5} \textrm{ or } x = 4$

Explanation:

Multiply both sides by (x-1) (x-3) :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$\frac{3}{x-1} \cdot (x-1) (x-3)+ \frac{4}{x- 3 } \cdot (x-1) (x-3) = 5\cdot (x-1) (x-3)$

$3 \cdot (x-3)+4\cdot (x-1) = 5\cdot (x-1) (x-3)$

$3x-9 + 4x-4 = 5 (x ^{2 } -4x+3)$

$7x-13= 5 x ^{2 } -20x+15$

$7x -13 -7x + 13= 5 x ^{2 } -20x -7x +15+ 13$

$5 x ^{2 } -27x + 28 = 0$

Factor this using the -method. We split the middle term using two integers whose sum is and whose product is $5 \cdot28 = 140$ . These integers are -20,-7 :

$5 x ^{2 } -20x -7x + 28 = 0$

$\left (5 x ^{2 } -20x \right ) - \left (7x - 2 8\right ) = 0$

$5x \left (x-4 \right ) - 7 \left (x-4 \right ) = 0$

$\left (5x - 7 \right )\left (x-4 \right ) = 0$

Set each factor equal to 0 and solve separately:

5x - 7 = 0

5x - 7 + 7 = 0 + 7

5x = 7

$5x \div 5 = 7 \div 5$

$x = \frac{7}{5} = 1 \frac{2}{5}$

x - 4 = 0

x - 4 + 4 = 0+ 4

x = 4

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Algebra 1 : Linear / Rational / Variable Equations

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #31 : Linear / Rational / Variable Equations

Example Question #152 : Equations / Inequalities

Example Question #32 : Linear / Rational / Variable Equations

Example Question #1 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #2 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #33 : Linear / Rational / Variable Equations

Example Question #1 : How To Find The Solution To A Rational Equation With Lcd

Example Question #2 : How To Find The Solution To A Rational Equation With Lcd

Example Question #1 : Solving Rational Expressions

Example Question #2 : Solving Rational Expressions

All Algebra 1 Resources

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