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Algebra 1 : How to write expressions and equations

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Example Question #1 : How To Write Expressions And Equations

Write in simplest form:

2 (4x + 5) - 3 (2x - 7)

Possible Answers:

2x + 31

2x + 12

2x - 2

2x - 11

2x + 3

Correct answer:

2x + 31

Explanation:

Rewrite, then distribute:

$\small \small 2 (4x + 5) - 3 (2x - 7)$

$\small = 2 (4x + 5) + (- 3) (2x - 7)$

$\small = 2 \cdot 4x + 2 \cdot 5 + (- 3) \cdot 2x - (-3) \cdot7$

$\small = \small 8x + 10 + (- 6) x - (-21)$

$\small \small = \small 8x + (- 6) x + 10 - (-21)$

$\small = 8x - 6 x + 10 + 21 = 2x + 31$

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Example Question #2 : How To Write Expressions And Equations

A car travels at a speed of 60 miles per hour. It is driven for 2.5 hours. How many miles does it travel?

Possible Answers:

120

150

180

100

Correct answer:

150

Explanation:

To solve this problem, you need to construct an algebraic equation. If is the distance traveled, then must equal to the speed multiplied by the time travelled. In this case, x=2.5*60 , which gives you a result of 150 miles.

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Example Question #3 : How To Write Expressions And Equations

Rewrite the expression in simplest terms, where is the imaginary number $\sqrt-1$ .

$i* \sqrt{-12}$

Possible Answers:

12i

$-2\sqrt3$

$-\sqrt{12}$

$2\sqrt3$

Correct answer:

$-2\sqrt3$

Explanation:

Writing this expression in simplest terms can be achieved by first factoring the radical into its smallest factors.

$i*\sqrt{-12}=\sqrt{-1} * (\sqrt{-1}\sqrt{4}\sqrt{3})$

Multiplying the two $\sqrt{-1}$ together results in . Multiplying this by $\sqrt4\sqrt3$ (which is simplified to $2\sqrt3$ ) results in the answer $-2\sqrt3$ .

$\sqrt{-1}*(\sqrt{-1}\sqrt{4}\sqrt{3})=(\sqrt{-1})^2*2\sqrt{3}=-2\sqrt{3}$

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Example Question #4 : How To Write Expressions And Equations

Rewrite the equation for in terms of .

$\frac{5x^5y}{12y^3} = \frac{4x}{7x^3y}$

Possible Answers:

$y = \frac{35x^7}{48}$

$x = \sqrt[7]{\frac{35y}{48}}$

$y = \frac{48}{35x^7}$

$x = \sqrt[7]{\frac{48y}{35}}$

Correct answer:

$y = \frac{35x^7}{48}$

Explanation:

The goal in expressing in terms of is to isolate on one side of the equation. One way to do this is to factor out of the fraction on the right side of the equation, then divide the entire equation by the fraction that remains after factoring. Remember that dividing by a fraction is the same as multiplying by the reciprocal of that fraction.

$\frac{5x^5y}{12y^3} = \frac{4x}{7x^3y} \rightarrow y (\frac{5x^5}{12y^3}) = \frac{4x}{7x^3y}$

$y (\frac{5x^5}{12y^3}) * (\frac{12y^3}{5x^5}) = \frac{4x}{7x^3y} * (\frac{12y^3}{5x^5})$

The left side of this equation will simply resolve into , although there are still variables on the right, so this is not yet in terms of . The right side resolves based on the rules for multiplying and dividing variables with exponents (add the exponents of like variables being multiplied, subtract the smaller exponent from the larger in the case of division, and change the variable to a if the resulting exponent is ).

$y = \frac{4x}{7x^3y} * (\frac{12y^3}{5x^5})$

$y = \frac{48xy^3}{35x^8y}$

$y = \frac{48y^2}{35x^7}$

Since there is still a y^2 in the numerator on the right side of the equation, we will need to divide both sides of the equation by y^2 .

$y = \frac{48y^2}{35x^7} \rightarrow (y) * \frac{1}{y^2}= (\frac{48y^2}{35x^7}) * \frac{1}{y^2}$

$\frac{y}{y^2}= \frac{1}{y} = \frac{48}{35x^7}$

We have no solved for the reciprocal of in terms of . We simply flip both sides of the equation to get our answer.

$y = \frac{35x^7}{48}$

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Example Question #1 : How To Write Expressions And Equations

Translate this sentence into a mathematical equation:

Three less than five times a number is the same as two more than twice that number.

Possible Answers:

5n-3 = 2n-2

5(n-3) = 2n+2

5n-3 = 2(n+2)

5n + 3 = 2n + 2

5n-3 = 2n+2

Correct answer:

5n-3 = 2n+2

Explanation:

Three less than five times a number is the same as two more than twice that number.

Let the number be .

"Three less than five times a number" translates into 5n-3 .

"Is the same as" means equal to or "".

"Two more than twice that number" means 2n+2 .

Putting these together gives:

5n-3 = 2n+2

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Example Question #6 : How To Write Expressions And Equations

For the given equation determine the slope:

3x-2y=5

Possible Answers:

$\frac{2}{3}$

$\frac{3}{2}$

Correct answer:

$\frac{3}{2}$

Explanation:

By changing the equation to slope intercept form we get the following:

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

Hence the slope is

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

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Example Question #7 : How To Write Expressions And Equations

What is the slope and the and intercepts of a line which passes through $\left ( -3,0 \right )$ and $\left ( -3,2 \right )$ ?

Possible Answers:

slope = 0, x-int = -3, y-int = 2

slope = 0, x-int = 2, y-int = -3

slope = undefined, x-int = -3, y-int = none

slope = 1, x-int = 2, y-int = 2

slope = undefined, x-int 2, y-int = -3

Correct answer:

slope = undefined, x-int = -3, y-int = none

Explanation:

For a vertical line e.g. x=-3 , run=0 and $\frac{rise}{0}=undefined$

This line does not intersect the y-axis and hence there is no y-intercept .

Since the line passes through $\left ( -3,0 \right )$ hence the -intercept =-3 .

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Example Question #1 : How To Write Expressions And Equations

Write the equation of a line with a slope of

$m=-\frac{3}{5}$

and passes through the point $\left ( 3,-2 \right )$ .

Possible Answers:

$\frac{3}{5}x +1$

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

$\frac{-3}{5}x-1$

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

$\frac{3}{5}x +1$

Correct answer:

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

Explanation:

Here we use the point-slope formula of a line which is

$\left ( y-y_{1} \right )=m\left ( x-x_{1} \right )$

By plugging in , $x_{1}$ , and $y_{1}$ values we get the following:

$\left ( y-\left ( -2 \right ) \right ) = \frac{-3}{5}\left ( x-5 \right )$

which is equal to

$\left ( y+2 \right ) =\frac{-3}{5}\left ( x-5 \right )$

When the above is simplified we get:

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

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Example Question #9 : How To Write Expressions And Equations

Complete the missing information for the equation of the following line

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

and determine which one of the $\left ( x,y \right )$ coordinates is not a solution to the above equation.

$\left ( 0,5 \right ),\left ( \frac{15}{2},0 \right ),\left ( -3,7 \right ),\left ( 3,3 \right ), \left ( -6,7 \right )$

Possible Answers:

$\left ( 3,3 \right )$

$\left ( 0,5 \right )$

$\left ( \frac{15}{2},0 \right )$

$\left ( -3,7 \right )$

$\left ( -6,7 \right )$

Correct answer:

$\left ( -6,7 \right )$

Explanation:

Replacing with , one gets y=9 which tells us that $\left ( -6,7 \right )$ is not a solution.

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Example Question #10 : How To Write Expressions And Equations

Convert the following into the standard form of a line:

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

Possible Answers:

3x+5y=10

5x-3y=10

3x-5y-10=0

5x+3y=10

3x-5y=10

Correct answer:

3x+5y=10

Explanation:

Multiplying each term of the given equation by the denominator of the slope which is 5 one gets :

5y=-3x+10

which can be written as

3x+5y=10

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Algebra 1 : How to write expressions and equations

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #1 : How To Write Expressions And Equations

Example Question #2 : How To Write Expressions And Equations

Example Question #3 : How To Write Expressions And Equations

Example Question #4 : How To Write Expressions And Equations

Example Question #1 : How To Write Expressions And Equations

Example Question #6 : How To Write Expressions And Equations

Example Question #7 : How To Write Expressions And Equations

Example Question #1 : How To Write Expressions And Equations

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