David W. answered 04/29/16
Tutor
4.7
(90)
Experienced Prof
a)
10x-8y=4 [eq a1]
-5x+3y=-9 [eq a2]
-5x+3y=-9 [eq a2]
To solve by elimination, we need one set of coefficients (either x-coefficients or y-coefficients) to be opposite or the same, so we can eliminate them (by addition or subtraction).
For this problem, let's pick the x term:
20x - 16y = 8 [multiply eq a1 by 2]
-20x +12y = -36 [multiply eq a2 by 4]
---------------------- [elimination; add equations]
-4y = -28
y = 7 [eq a3]
8y = 56 [multiply eq a3 by 8]
10x - 8y = 4 [eq a1 again]
--------------------- [elimination; add equations]
10x = 60
x = 6
Checking (very important):
Is 10*6 -8*7 = 4 ?
60 - 56 = 4 ?
4 = 4 ?yes
Is (-5)*6 + 3*7 = -9 ?
-30 + 21 = -9 ?
-9 = -9 ?yes
b)
-3x-10y= 4 [eq b1]
x- 5y=18 [eq b2]
x- 5y=18 [eq b2]
Let's eliminate y this time --
-3x -10y = 4 [eq b1 again]
-2x +10y = -36 [multiply eq b2 by (-2)]
-------------------- [elimination; add equations]
-5x = -32
x = 32/5
Special note: You may proceed and solve for y, but I think that there was a typo and that equation b1 should have been: -3x -10y =-4. I will solve that system:
-3x -10y = -4
-2x +10y = -36
---------------------
-5x = -40
x = 8
x -5y=18 [elimination; subtract this eq from the line above]
-----------------
5y = -10
y = -2
Now, you should check that answer (8,-2) and double check whether there was a typo. If no typo, then you know how to solve by elimination, right?