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SAT II Math I : Solving Linear Functions

Study concepts, example questions & explanations for SAT II Math I

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All SAT II Math I Resources

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Example Questions

Example Question #1 : Solving Linear Functions

Solve for in terms of .

3y-xy+7=2

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

First, subtract 7 to both sides.

3y-xy+7=2

3y-xy=-5

Factor the y on the left hand side.

y(3-x)=-5

Divide both sides by 3 – x.

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

Take an extra step by factoring the minus sign on the denominator.

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

Cancel the minus signs.

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

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Example Question #1 : Solving Linear Functions

Find the equation for the line goes through the two points below.

(1,-2)

(-3, 6)

Possible Answers:

y = 2x -4

y = -2x

y = 4x - 2

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

$y = \frac{1}{2}x + 4$

Correct answer:

y = -2x

Explanation:

Let $P_{1} = (1, \; -2) \; \; and \; \; P_{2} = (-3, \; 6)$ .

First, calculate the slope between the two points.

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{6 - (-2)}{-3 - 1} = \frac{8}{-4} = -2$

Next, use the slope-intercept form to calculate the intercept. We are able to plug in our value for the slope, as well the the values for P_1 .

y = mx + b

-2 = -2(1) + b

b=0

Using slope-intercept form, where we know m=-2 and b=0 , we can see that the equation for this line is y = -2x .

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Example Question #1 : Solving Linear Functions

Find the equation of the line passing through the points $\small (-3,2)$ and $\small (1,8)$ .

Possible Answers:

$\small y=-\frac{3}{2}x+\frac{13}{2}$

$\small y=\frac{3}{2}x+\frac{13}{2}$

$\small y=\frac{2}{3}x+\frac{13}{2}$

$\small y=\frac{2}{3}x-\frac{13}{2}$

$\small y=-\frac{2}{3}x+\frac{13}{2}$

Correct answer:

$\small y=\frac{3}{2}x+\frac{13}{2}$

Explanation:

To calculate a line passing through two points, we first need to calculate the slope, $\small m$ .

$\small m=\frac{\Delta y}{\Delta x} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{8-2}{1-(-3)} = \frac{6}{4} = \frac{3}{2}$

Now that we have the slope, we can plug it into our equation for a line in slope intercept form.

$\small y=\frac{3}{2}x+b$

To solve for $\small b$ , we can plug in one of the points we were given. For the sake of this example, let's use $\small (-3,2)$ , but realize either point will give use the same answer.

$\small 2=\frac{3}{2}(-3)+b$

$\small 2=\frac{-9}{2}+b$

$\small 2+\frac{9}{2} = b$

$\small \frac{13}{2} = b$

Now that we have solved for b, we can plug that into our slope intercept form and produce and the answer

$\small y=\frac{3}{2}x+\frac{13}{2}$

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Example Question #1 : Solving Linear Functions

Find the point at which these two lines intersect:

$\small 2x+5y=8$

$\small \small \small -2x+2y=-1$

Possible Answers:

$\left(\frac{3}{2},1\right)$

$\left(1,-\frac{3}{2}\right)$

$\left(-\frac{3}{2},1\right)$

$\left(1,\frac{3}{2}\right)$

Correct answer:

$\left(\frac{3}{2},1\right)$

Explanation:

We are looking for a point, $\small (x,y)$ , where these two lines intersect. While there are many ways to solve for $\small x$ and $\small y$ given two equations, the simplest way I see is to use the elimination method since by adding the two equations together, we can eliminate the $\small x$ variable.

7y=7

Dividing both sides by 7, we isolate y.

$\small y=1$

Now, we can plug y back into either equation and solve for x.

$\small 2x+(5)(1)=8$

$\small 2x+5=8$

Next, we can solve for x.

$\small 2x=3$

$\small x=(3/2)$

Therefore, the point where these two lines intersect is $\left(\frac{3}{2}, 1\right)$ .

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Example Question #5 : Solving Linear Functions

$y=\frac{1}{3}x+9$

Solve for when y=2 .

Possible Answers:

Correct answer:

Explanation:

The first thing is to plug in the given value so you equation is

$2=\frac{1}{3}x+9$ .

Then you must subtract from both sides in order to get by itself.

You now have

$-7=\frac{1}{3}x$

and you must multiply by for each side and you get x=-21 .

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Example Question #1 : Solving Linear Functions

Solve: -7x-16 = 3x+2

Possible Answers:

$-\frac{9}{5}$

$\frac{3}{4}$

$-\frac{7}{2}$

$-\frac{9}{2}$

Correct answer:

$-\frac{9}{5}$

Explanation:

Add on both sides.

-7x-16 +(7x)= 3x+2+(7x)

-16 = 10x+2

Subtract 2 from both sides.

-16 -2= 10x+2-2

-18 = 10x

Divide by 10 on both sides.

$\frac{-18}{10} =\frac{ 10x}{10}$

Reduce the fractions.

The answer is: $-\frac{9}{5}$

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Example Question #1 : Solving Linear Functions

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

Possible Answers:

$\frac{4}{3}$

$\frac{7}{3}$

$\frac{1}{3}$

Correct answer:

Explanation:

Distribute the right side.

-2(2-3x) = -4 +6x

Rewrite the equation.

3x+5 = -4+6x

Subtract on both sides.

3x+5 -3x= -4+6x-3x

5=-4+3x

Add 4 on both sides.

5+4=-4+3x+4

9=3x

Divide by three on both sides.

$\frac{9}{3}=\frac{3x}{3}$

The answer is:

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SAT II Math I : Solving Linear Functions

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #1 : Solving Linear Functions

Example Question #1 : Solving Linear Functions

Example Question #1 : Solving Linear Functions

Example Question #1 : Solving Linear Functions

Example Question #5 : Solving Linear Functions

Example Question #1 : Solving Linear Functions

Example Question #1 : Solving Linear Functions

All SAT II Math I Resources

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