Exercise 133 Page 568

Consider the given equation of a line. 3x+2y=10 When lines are parallel they have the same slope. To help us identify the slope of this line, let's first convert it into slope-intercept form, y=mx+ b, where m is the slope and (0, b) is the y-intercept.

3x+2y=10

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Solve for y

SubEqn

LHS-3x=RHS-3x

2y=-3x+10

DivEqn

.LHS /2.=.RHS /2.

y=-3/2x+5

MoveNegNumToFrac

Put minus sign in front of fraction

y=-3/2x+5

With this information we can more easily identify the slope m and y-intercept b. y=-3/2x+ 5 Now we know that all lines that are parallel to the line whose equation is given will have a slope of - 32. Thus, we can write a general equation in slope-intercept form for these lines. y=- 3/2x+ b We are asked to write the equation of a line parallel to the one with given equation that passes through the point ( 4, - 7). By substituting this point into the above equation for x and y, we will be able to solve for the y-intercept b of the parallel line.

y=- 3/2x+b

SubstituteII

x= 4, y= - 7

- 7=- 3/2( 4)+b

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Solve for b

MoveRightFacToNum

a/c* b = a* b/c

- 7=- 12/2+b

CalcQuot

Calculate quotient

- 7=- 6+b

AddEqn

LHS+6=RHS+6

- 1=b

RearrangeEqn

Rearrange equation

b=- 1

Now that we have the y-intercept, we can write the parallel line to y=- 32x+5 that goes through (4,- 7). y=- 3/2x+( - 1) ⇔ y=- 3/2x-1