SOLUTION: Solve by elimination method
1. x+4y=16
3x+5y=20
2. Bx+4y=16
4x-7y=26
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Question 1197248: Solve by elimination method
1. x+4y=16
3x+5y=20
2. Bx+4y=16
4x-7y=26
Found 2 solutions by josgarithmetic, math_tutor2020:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
#2:
----------------
Bx+4y=16
4x-7y=26
----------------
7*E1 and 4*E2
--------------
7Bx+28y=112
16x-28y=104
-------------
ADD corresponding terms.
Back to the original system,
4*E1 and B*E2
---------------
4Bx+16y=64
4Bx-7By=26B
--------------
E2-E1
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
I'll do problem 1 to get you started.
The given system is this
x+4y = 16
3x+5y = 20
Triple everything in the first equation to go from
x+4y = 16
to
3x+12y = 48
This is done so that the x terms have the same coefficient.
Here's what the system of equations looks like now
3x+12y = 48
3x+5y = 20
Now subtract the equations straight down
3x-3x turns into 0x or 0, so the x terms go away
12y-5y turns into 7y
48-20 turns into 28
We get 0x+7y = 28 or 7y = 28
Divide both sides by 7 to get y = 4
Now use this y value to find x.
We can use it with any equation involving x and y mentioned earlier.
I'll use the first equation
x+4y = 16
x+4*4 = 16
x+16 = 16
x = 16-16
x = 0
Answer: The solution is (x,y) = (0,4)
Check:
Plug x = 0 and y = 4 into the first equation
x+4y = 16
0+4*4 = 16
16 = 16
That works out. Now do the same for the second equation
3x+5y = 20
3*0+5*4 = 20
20 = 20
That is confirmed as well.
Both equations are true for x = 0 and y = 4, which fully confirms the solution.
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