Question

at what point does a line with a slope of 3/4 and a y- intercept of -5 intersect a line with a slope of -1/4 and a y- intercept of 3?

Answer 1

Answer:

(8,1)

Step-by-step explanation:

I don't know how you are supposed to answer this, there are several methods.  I will use the earliest algebraic method I know, if you were supposed to do this graphically then you should graph these lines on graph paper and see where they cross

line 1: y = (3/4)x - 5

line 2: y = (-1/4)x + 3

at the point these lines cross the x and y values of the two lines are the same, so you can write

y (line 1) = y (line 2) or

(3/4)x - 5 = (-1/4)x + 3, subtract 3 from both sides

(3/4)x - 8 = (-1/4)x, multiply both sides by 4 (to eliminate the denominators)

3x -32 = -x, add x, add 32 to both sides

4x = 32, divide both sides by 4

x = 8, substitute x = 8 into either of the linear equations

y=(3/4)x - 5 = (3/4)(8) - 5 = (24/4) - 5 = 6 - 5 = 1

so the intersection point is (8,1)

Given that the answer worked out well, the graphic method should be good too.

Answer 2

Answer:

The equation of a line in slope-intercept form is y = mx + b

where m is the slope

and b is the y-intercept.

For example,

   in the equation  y = 9x - 7

   the slope is 9

   and the y-intercept is -7, which means that the line crosses the y-axis at the point (0, -7)

If your question you are given m=3/8 and you can find the value of b from the given y-intercept.

Step-by-step explanation:

The slope-intercept form of a linear equation is:

y

=

m

x

+

b

Where

m

is the slope and

b

is the y-intercept value.

Substituting gives:

y

=

3

x

+

7