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Algebra 1 : Systems of Equations

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Example Questions

Example Question #1 : Systems Of Equations

Solve for :

8x-5-4x=-x+10

Possible Answers:

Correct answer:

Explanation:

8x-5-4x=-x+10 can be simplified to become

4x-5=-x+10

Then, you can further simplify by adding 5 and to both sides to get 5x=15 .

Then, you can divide both sides by 5 to get x=3 .

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Example Question #1 : How To Find The Solution To An Equation

What number is six less than eight more than one half of four times the square of the greatest negative integer?

Possible Answers:

$\frac{1}{3}$

Correct answer:

Explanation:

Turn the word problem into math. Start at the end--what is the greatest negative integer? -1!

$(-1)^2\cdot4\cdot\frac{1}{2}+8-6=1\cdot4\cdot\frac{1}{2}+8-6=2+8-6=4$

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Example Question #2 : How To Find The Solution To An Equation

$18\div\frac{1}{6}\cdot\frac{9}{72}\cdot\frac{4}{27}\cdot(-8+7)^9=$

Possible Answers:

Correct answer:

Explanation:

$18\div\frac{1}{6}\cdot\frac{9}{72}\cdot\frac{4}{27}\cdot(-8+7)^9=18\cdot6\cdot\frac{1}{8}\cdot\frac{4}{27}\cdot-1$

Remember to cancel as you go--the 18 will cancel with the 27, the 4 will cancel with the 8, and so on:

$18\cdot6\cdot\frac{1}{8}\cdot\frac{4}{27}\cdot-1= 2\cdot6\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot-1$

Continue to reduce:

$2\cdot6\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot-1= 2\cdot1\cdot-1=-2$

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Example Question #1 : Linear Equations

Solve for :

6x-1=12x+8-3x

Possible Answers:

Correct answer:

Explanation:

To solve for , you must first combine the 's on the right side of the equation. This will give you $\ 6x-1=9x+8$ .

Then, subtract and from both sides of the equation to get -9=3x .

Finally, divide both sides by to get the solution x=-3 .

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Example Question #2 : How To Find The Solution To An Equation

Solve for :

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

Possible Answers:

Correct answer:

Explanation:

First, use the distributive property to simplify the right side of the equation: 3x+5=6x-4 . Then, subtract and add 4 to both sides to separate the 's and the integers to get 9=3x . Divide both sides by 3 to get x=3 .

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Example Question #3 : How To Find The Solution To An Equation

Solve for :

a(x+b)= c

Possible Answers:

$\frac{a}{c-b}$

$\frac{c-b}{a}$

$\frac{a-b}{c}$

$\frac{c}{a}-b$

$\frac{a}{c}-b$

Correct answer:

$\frac{c}{a}-b$

Explanation:

To solve for , first divide both sides by : $x+b=\frac{c}{a}$ . Then, subtract both sides by to get $x=\frac{c}{a}-b$ .

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Example Question #3 : How To Find The Solution To An Equation

(4^2)^x=64

Possible Answers:

$\frac{3}{2}$

$-\frac{3}{4}$

Correct answer:

$\frac{3}{2}$

Explanation:

$(4^2)^x= 4^{2x}=64$

$64= 4\cdot 4\cdot 4=4^3$

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

2x = 3

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

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Example Question #4 : How To Find The Solution To An Equation

$\frac{4^2+6\cdot 9}{4+3(2)}+\frac{7}{14}\cdot\frac{50}{25}-1=$

Possible Answers:

19.8

Correct answer:

Explanation:

1. First simplify the first expression:

$\frac{4^2 +6\cdot 9}{4+3(2)}=\frac{4\cdot 4+54}{4+6}=\frac{16+54}{10}=\frac{70}{10}=7$

2. Then, simplify the next two expressions:

$\frac{7}{14}\cdot\frac{50}{25}=\frac{1}{2}\cdot\frac{2}{1}=1$

3. Finally, add and subtract:

7+1-1=7

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Example Question #3 : How To Find The Solution To An Equation

Solve for x.

$\frac{x^2-x^3-(-x-4)}{\frac{1}{x}+\frac{x}{1}}, x=-2$

Possible Answers:

$\frac{28}{3}$

$\frac{-28}{5}$

$\frac{-7}{9}$

$\frac{-4}{3}$

Correct answer:

$\frac{-28}{5}$

Explanation:

1. First solve for the numerator by plugging in -2 for x:

(-2)^2-(-2)^3-(-(-2)-4)=4+8+2= 14

2. Then, solve the denominator by combining the fractions:

$\frac{1}{x}+\frac{x}{1}= \frac{-1}{2}+\frac{-2}{1}=\frac{-1}{2}+\frac{-4}{2}=\frac{-5}{2}$

3. Finally, "rationalize" the complex fraction by multiplying top and bottom by -2/5:

$\frac{14}{\frac{-5}{2}}\cdot\frac{\frac{-2}{5}}{\frac{-2}{5}}=\frac{\frac{-28}{5}}{1}=\frac{-28}{5}$

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Example Question #6 : How To Find The Solution To An Equation

$\frac{x*7+5*4+1}{y*2+4}=3$

If x/y is equivalent to 12/20, what is the value of x?

Possible Answers:

Correct answer:

Explanation:

$\frac{x*7 +5*4+1}{y*2+4}= \frac{7x +21}{2y +4}=3$

Multiply both sides by the denominator (2y +4) to cancel it:

$\frac{7x +21}{2y +4}*(2y +4)=3*(2y+4)$

7x + 21=6y +12

Now, use substitution to solve for x:

$\frac{x}{y}=\frac{12}{20}-->20x=12y-->10x=6y$

Substitute 10x for 6y in the first equation:

7x+21=(10x)+12

9=3x

3=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 #1 : Systems Of Equations

Example Question #1 : How To Find The Solution To An Equation

Example Question #2 : How To Find The Solution To An Equation

Example Question #1 : Linear Equations

Example Question #2 : How To Find The Solution To An Equation

Example Question #3 : How To Find The Solution To An Equation

Example Question #3 : How To Find The Solution To An Equation

Example Question #4 : How To Find The Solution To An Equation

Example Question #3 : How To Find The Solution To An Equation

Example Question #6 : How To Find The Solution To An Equation

All Algebra 1 Resources

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