SOLUTION: Solve the equations using elimination and substitution (1)2x+3y=11 4x+y=12 (2)3x+2y=24 2x-3y=42

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Question 1134702: Solve the equations using elimination and substitution
(1)2x+3y=11
4x+y=12
(2)3x+2y=24
2x-3y=42

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

2x%2B3y=11.....eq.1
4x%2By=12......eq.2
------------------------
start with
4x%2By=12......eq.2...solve for y
y=-4x%2B12.......eq.2a...substitute in eq.1 to eliminate variable y


2x%2B3%28-4x%2B12%29=11.....eq.1
2x-12x%2B36=11...solve for x
-10x%2B36=11
-11%2B36=10x
25=10x
25%2F10=x
x=5%2F2
go back to eq.2a, substitute 5%2F2 for x
y=-4%2A%285%2F2%29%2B12
y=-10%2B12
y=2

solution is intersection point (x,y)=(5%2F2,2)


3x%2B2y=24.......eq.1
2x-3y=42.......eq.2
---------------------------
start with
3x%2B2y=24.......eq.1 ..solve for y
2y=-3x%2B24
y=-3x%2F2%2B24%2F2
y=-3x%2F2%2B12...........eq.1a
go to
2x-3y=42.......eq.2...substitute in y from eq.1a to eliminate variable y
2x-3%28-3x%2F2%2B12%29=42
2x%2B9x%2F2-36=42
2x%2B9x%2F2=42%2B36.........both sides multiply by 2
4x%2B9x=84%2B72
13x=156
x=156%2F13
x=12
go back to eq.1a
y=-3x%2F2%2B12...........eq.1a substitute 12 for x and solve for y
y=-%283%2A12%29%2F2%2B12
y=-%283%2Across%2812%296%29%2Fcross%282%29%2B12
y=-3%2A6%2B12
y=-18%2B12
y=-6

solution is intersection point (x,y)=(12,-6)