How do you solve the system of equations #3y-4x=-33# and #5x-3y=40.5#?

1 Answer
Jul 3, 2018

Answer:

See a solution process below:

Explanation:

Step 1) Solve each equation for #3y#:

  • Equation 1:

#3y - 4x = -33#

#3y - 4x + color(red)(4x) = -33 + color(red)(4x)#

#3y - 0 = -33 + 4x#

#3y = -33 + 4x#

  • Equation 2:

#5x - 3y = 40.5#

#5x - color(red)(5x) - 3y = 40.5 - color(red)(5x)#

#0 - 3y = 40.5 - 5x#

#-3y = 40.5 - 5x#

#color(red)(-1) xx -3y = color(red)(-1)(40.5 - 5x)#

#3y = (color(red)(-1) xx 40.5) - (color(red)(-1) xx 5x)#

#3y = -40.5 - (-5x)#

#3y = -40.5 + 5x#

Step 2) Because the left side of both equations are equal we can equate the right side of both equations and solve for #x#:

#-33 + 4x = -40.5 + 5x#

#-33 + color(blue)(40.5) + 4x - color(red)(4x) = -40.5 + color(blue)(40.5) + 5x - color(red)(4x)#

#7.5 + 0 = 0 + (5 - color(red)(4))x#

#7.5 = 1x#

#7.5 = x#

#x = 7.5#

Step 3) Substitute #7.5# for #x# into either of the solutions for the equations in Step 1 and calculate #y#:

#3y = -33 + 4x# becomes:

#3y = -33 + (4 xx 7.5)#

#3y = -33 + 30#

#3y = -3#

#(3y)/color(red)(3) = -3/color(red)(3)#

#(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = -1#

#y = -1#

The Solution Is:

#x = 7.5# and #y = -1#

Or

#(7.5, -1)#