How do you solve this system of equations by graphing: #-5x + 4y = 3# and #x = 2y - 15#?

2 Answers
iceman
Feb 1, 2018

#(x,y)=(9,12)#

Explanation:

Okay so we are solving a linear system of equations consist of two straight lines:

#-5x+4y=3#
#x=2y-15#
This is an ideal set up for "Substitution" method, so we substitute x in the first equation and solve for y:
#-5(2y-15)+4y=3#
#-10y+75+4y=3#
#-6y=-72#
#y=12#
now we plug this in one of the equations and solve for x:
#x=2*12-15#
#x=24-15#
#x=9#
thus:
#(x,y)=(9,12)#

smendyka
Feb 2, 2018

I’m trying to get the answer to graph it’s -5x+4y=3 x=2y-15

Explanation:

To solve this through graphing you must find two points for each equation, draw a line through each set of points and then determine where they intersect:

Equation 1:

  • For: #x = -3#
    #(-5 xx -3) + 4y = 3#
    #15 + 4y = 3#
    #-color(red)(15) + 15 + 4y = -color(red)(15) + 3#
    #0 + 4y = -12#
    #4y = -12#
    #(4y)/color(red)(4) = -12/color(red)(4)#
    #y = -3# or #(-3, -3)

  • For: #x = 1#
    #(-5 xx 1) + 4y = 3#
    #-5 + 4y = 3#
    #color(red)(5) - 5 + 4y = color(red)(5) + 3#
    #0 + 4y = 8#
    #4y = 8#
    #(4y)/color(red)(4) = 8/color(red)(4)#
    #y = 2# or #(1, 2)

  • Plot The Two Points and Draw The Line:

graph{(-5x+4y-3)((x+3)^2+(y+3)^2-0.25)((x-1)^2+(y-2)^2-0.25)=0 [-30, 30, -15, 15]}

Equation 2:

  • For: #y = 5#
    #x = (2 xx 5) - 15#
    #x = 10 - 15#
    #x = -5# or #(-5, 5)#

  • For: #y = 7#
    #x = (2 xx 7) - 15#
    #x = 14 - 15#
    #x = -1# or #(-1, 7)#

  • Plot The Two Points and Draw The Second Line:

graph{(x-2y+15)(-5x+4y-3)((x+5)^2+(y-5)^2-0.25)((x+1)^2+(y-7)^2-0.25)=0 [-30, 30, -15, 15]}

We can see the lines intersect at:

graph{(x-2y+15)(-5x+4y-3)((x-9)^2+(y-12)^2-0.075)=0 [-10, 20, -2, 13]}

#x = 9# and #y = 12# Or #(9, 12)#

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