Algebra

Can you please help me with this problem:

The x-intercept of a line has a value of -7/6 and a y-intercept of 7. Another line has a slope that is twice the slope of the first line an a y-intercept that is 15 more than the first lines' y-intercept. What is the product of the coordinates in the solution to this system? Write a system of equations for this situation and find its solution.

This is what I have:

x = -7/6 which is -1.16 - twice this amount is 2.33

y = 7 (15 more than that first y intercept is 22)

coordinates are 2.33 and 22...is this correct? Thank you

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asked by Chris
  1. First of all , don't switch to decimals
    ( 7/6 ≠ 1.16, it is 1.1666.... )

    So you have two points (-7/6 , 0) and (0,7)
    slope = (7-0)/(0+7/6)
    = 7/(7/6) = 6
    equation of first line : y = 6x + 7

    slope of 2nd line = 12 <----- it said twice the first
    y-intercept of 2nd line = 7+15 = 23
    so for the second line
    equation is y = 12x + 23

    solve them:
    12x+23 = 6x+7
    6x = -16
    x = -16/6 = -8/3
    then y = 6(-8/3) + 7 = -9

    so the product of the coordinates = (-8/3)(-9)
    = 24

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    posted by Reiny

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