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Algebra 1 : How to find the solution to an equation

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Example Questions

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

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 #2 : Systems Of Equations

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 #3 : Systems Of Equations

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

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 #4 : Systems Of 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 #1 : 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 #1 : Systems Of Equations

Solve for :

a(x+b)= c

Possible Answers:

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

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

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

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

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

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 #2 : How To Find The Solution To An Equation

(4^2)^x=64

Possible Answers:

$-\frac{3}{4}$

$\frac{3}{2}$

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 #1 : 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 #1 : 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:

$\frac{-4}{3}$

$\frac{-28}{5}$

$\frac{-7}{9}$

$\frac{28}{3}$

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 #1 : 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 : How to find the solution to an equation

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

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

Example Question #2 : Systems Of Equations

Example Question #3 : Systems Of Equations

Example Question #4 : Systems Of Equations

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

Example Question #1 : Systems Of Equations

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

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

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

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

All Algebra 1 Resources

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