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Algebra 1 : Equations / Solution Sets

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Example Questions

Example Question #21 : Equations / Solution Sets

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 #2 : Solving 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 #22 : Equations / Solution Sets

Solve for :

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

Possible Answers:

x = -25

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

The equation has no solution.

x = 225

x = 125

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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Example Question #42 : Basic Single Variable Algebra

We have three cats, Chai, Sora, and Newton. Chai is 3 years old. Sora two years older than twice Chai's age. Newton is one year younger than one-fourth of Sora's age. How old are Sora and Newton?

Possible Answers:

Sora: 8 years

Newton: 1 year

Sora: 3 years

Newton: not born yet

Sora: 4 years

Newton: one half year

Sora: 1 year

Newton: 8 years

Sora: 5 years

Newton: 6 years

Correct answer:

Sora: 8 years

Newton: 1 year

Explanation:

To make this much easier, translate the word problem into a system of three equations.

C=3

S=2C+2

$N=\frac{1}{4}S-1$

We have C for Chai, S for Sora, and N for Newton. To find Sora's age, plug in into .

S=2(3)+2=8

Sora is 8 years old. Use this to find Newton's age.

$N=\frac{1}{4}(8)-1=1$

Newton is one year old. So the answer is:

Sora, 8 years

Newton, 1 year

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Example Question #1 : Solutions And Solution Sets

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Possible Answers:

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

$\left \{ 4,9\right \}$

$\left \{ -9,-4,4,9\right \}$

The equation has no real solutions.

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

Correct answer:

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

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:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as (u+?)(u+?) , replacing the two question marks with integers with product 36 and sum ; these integers are -9,-4 .

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

Substitute back:

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

These factors can themselves be factored as the difference of squares:

(x+3)(x-3)(x+2)(x-2) = 0

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

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

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

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Example Question #111 : Solving Rational Expressions

Solve for :

$\frac{6}{x-2} + 1 = x$

Possible Answers:

$x = -1 \textrm{ or } x = 4$

There is no solution.

$x = 2 \textrm{ or } x = 4$

x = -1

$x = 1 \textrm{ or } x = -4$

Correct answer:

$x = -1 \textrm{ or } x = 4$

Explanation:

Subtract 1 from both sides, then multiply all sides by x-2 :

$\frac{6}{x-2} + 1 = x$

$\frac{6}{x-2} + 1 - 1 = x- 1$

$\frac{6}{x-2} = x- 1$

$\frac{6}{x-2} \cdot (x-2) = \left (x- 1 \right ) \cdot (x-2)$

$6 = x^{2} -3x + 2$

$6 - 6 = x^{2} -3x + 2 - 6$

$0 = x^{2} -3x -4$

A quadratic equation is yielded. We can factor the expression, then set each individual factor to 0.

$x^{2} -3x -4 =0$

(x-4) (x+1) = 0

$x-4 = 0 \Rightarrow x = 4$

$x+1 = 0 \Rightarrow x = -1$

Both of these solutions can be confirmed by substitution.

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Example Question #141 : Systems Of Equations

Solve for :

$x = \sqrt{2x-1} +8$

Possible Answers:

$x = 5 \textrm{ or } x = 13$

$x = -5 \textrm{ or } x = 5$

$x = -13 \textrm{ or } x = 13$

x = 13

x = 5

Correct answer:

x = 13

Explanation:

Isolate the radical, square both sides, and solve the resulting quadratic equation.

$x = \sqrt{2x-1} +8$

$x - 8 = \sqrt{2x-1} +8 -8$

$x - 8 = \sqrt{2x-1}$

$\left (x -8 \right )^{2} =\left ( \sqrt{2x-1} \right )^{2}$

$x^{2} - 2 \cdot x \cdot 8 + 8^{2} = 2x - 1$

$x^{2} - 16 x + 64 = 2x - 1$

$x^{2} - 16 x + 64 - 2x + 1 = 2x - 1 - 2x + 1$

$x^{2} - 18 x + 65 = 0$

Factor the expression at left by finding two integers whose product is 65 and whose sum is ; they are -13,-5 .

(x-13) (x-5 ) = 0

Set each linear binomial to 0 and solve for to find the possible solutions.

x-13 = 0

x = 13

x-5 = 0

x = 5

Substitute each for .

x=5:

$x = \sqrt{2x-1} +8$

$5 = \sqrt{2 \cdot 5-1} +8$

$5 = \sqrt{9} +8$

5 = 3 +8

5 = 11

This is a false statement, so 5 is a false "solution".

x=13:

$x = \sqrt{2x-1} +8$

$13 = \sqrt{2 \cdot 13-1} +8$

$13= \sqrt{25} +8$

13 = 5 +8

13 = 13

This is a true statement, so 13 is the only solution of the equation.

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Example Question #261 : Equations / Inequalities

Solve the equation for .

$\left | 5x-4\right |=\left | 2x+7\right |$

Possible Answers:

$x=\frac{11}{3}\;or\;x=-\frac{3}{7}$

$x=\frac{3}{7}\:\: or\: \:x=-\frac{11}{3}$

$x=\frac{11}{7}\;or\;x=-\frac{11}{7}$

$x=0\;or\;x=-2$

$x=\frac{5}{8}\;or\;x=4$

Correct answer:

$x=\frac{11}{3}\;or\;x=-\frac{3}{7}$

Explanation:

The two sides of the equation will be equal if the quantities inside the absolute value signs are equal or equal with opposite signs.

From this, you get two equations.

5x-4=2x+7 or 5x-4=-(2x+7)

3x=11 7x=-3

$x=\frac{11}{3}$ $x=-\frac{3}{7}$

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Example Question #266 : Equations / Inequalities

Find the solution, in the form (x,y) , to the following system of equations:

4x+3y=9

-3x-5y=-26

Possible Answers:

(-3,7)

(3,-1)

(-1,4)

(1,-3)

(3,-7)

Correct answer:

(-3,7)

Explanation:

(1) $ $ $ $ 4x+3y=9

(2)$ $ $ $ -3x-5y=-26

Multiply (1) by 3, multiply (2) by 4:

$3\cdot (4x+3y=9)=12x+9y=27$

$4\cdot(-3x-5y=-26)=-12x-20y=-104$

Add the two resulting solutions:

(12x+9y)+(-12x-20y)=27-104

-11y=-77

y=7

Substitute y=7 into (1) and solve for :

4x+3(7)=9

4x+21-21=9-21

4x=-12

x=-3

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Example Question #267 : Equations / Inequalities

2x-7=11

Possible Answers:

x=5

x=2

x=9

None of the other answers.

No solution.

Correct answer:

x=9

Explanation:

To find the solution isolate the variable on one side of the equation.

2x-7=11

2x=18

x=9

To check to see if this is the correct solution, plug the value of x back into the equation and solve:

2(9)-7=11

18-7=11

11=11

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Algebra 1 : Equations / Solution Sets

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #21 : Equations / Solution Sets

Example Question #2 : Solving Equations

Example Question #22 : Equations / Solution Sets

Example Question #42 : Basic Single Variable Algebra

Example Question #1 : Solutions And Solution Sets

Example Question #111 : Solving Rational Expressions

Example Question #141 : Systems Of Equations

Example Question #261 : Equations / Inequalities

Example Question #266 : Equations / Inequalities

Example Question #267 : Equations / Inequalities

All Algebra 1 Resources

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