Question 1159946: solving system of equations by substitution y=1/3x-4 and y=-7/3x+4? Found 2 solutions by Theo, MathTherapy:Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! y = 1/3 * x - 4
y = -7/3 * x + 4
in the second equation, replace y with 1/3 * x - 4 to get:
1/3 * x - 4 = -7/3 * x + 4
add 4 to both sides of the equation to get:
1/3 * x = -7/3 * x + 4 + 4
combine like terms to get:
1/3 * x = -7/3 * x + 8
add 7/3 * x to both sides of the equation to get:
1/3 * x + 7/3 * x = 8
combine like terms to get:
8/3 * x = 8
multiply both sides of the equation by 3 to get:
8 * x = 8 * 3
divide both sides of the equation by 8 to get:
x = 3
x = 3 should be your solution.
when x = 3:
y = 1/3 * x - 4 becomes y = 1/3 * 3 - 4 which becomes y = 1 - 4 which becomes y = -3.
and:
y = -7/3 * x + 4 becomes y = -7/3 * 3 + 4 which becomes y = -7 + 4 which becomes y = -3.
both equations are solved simultaneously when the value of x = 3.
y = -3 in both equations when the value of x = 3.
these equations were solved by substitution because you substituted the value of y in the second equation by its equivalent from the first equation.
We can either substitute eq (i) into eq (ii), or vice-versa, which ultimately will result in the equations being set equal to each other, as in:
x - 12 = - 7x + 12 ------ Multiplying by LCD, 3
x + 7x = 12 + 12
8x = 24 ----- Substituting 3 for x in eq (i)
