SOLUTION: How would I solve the equation listed below using the elimination method? I have tried, but I keep coming up with differnt answers.
x+5y=29
-7x+4y=70
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Question 208388: How would I solve the equation listed below using the elimination method? I have tried, but I keep coming up with differnt answers.
x+5y=29
-7x+4y=70
Found 2 solutions by nerdybill, jadi929:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
x+5y=29
-7x+4y=70
.
Multiply the top equation by 7 then combine:
7x+35y=203
-7x+4y=70
----------
39y = 273
y = 7
.
Substitute the above back into equation 1 to solve for x:
x+5y=29
x+5(7)=29
x+35=29
x = 29-35
x = -6
.
Therefore, the solution is:
(x,y) = (-6, 7)
Answer by jadi929(1) (Show Source): You can put this solution on YOUR website!
So you have this problem:
x + 5y = 29
-7x + 4y = 70
You want to multiply everything in the first equation by positive 7, because that would give you 7x as the first term in the first equation, and will cancel out with -7 in the second equation.
So you multiply 7 thoughout the first equation:
7x+35y=203 < first equation
-7x+4y=70
The two 7s cancel out, and you add the other stuff.
You are left with:
39y = 273
divide 273 by 39, and y = 7
Next you take the one of the equations and plug 7 in for y and find x.
I chose the second equation:
-7x + 4y = 70
-7x+ 4(7) = 70
-7x + 28 = 70 subtract 28 from 70
-7x=42
x= -6
Answer is (-6, 7)
You can plug it back into both original equations to verify if its correct, which i did.
-7(-6) + 4(7) = 70
42+28= 70
70=70
x+5y=29
-6+5(7)=29
-6+35=29
29=29
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