Solve the following by elimination method: 2x - 3y = 5 3x - (2y-3)/5 = 4
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calendarEducator since 2011
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(1) 2x-3y=5
(2) 3x-(2y-3)/5=4
Simplify (2)
(5)3x - (5)(2y-3)/5 = 4(5)
15x - 2y + 3 = 20
(3) 15x - 2y = 17
2x - 3y = 5 multiply by -2 to get -4x + 6y = -10 (4)
15x - 2y = 23 multiply by 3 to get 45x - 6y = 51 (5)
Now add (4) and (5) to get
41x = 41 and solving for x we get
x = 1
To find y we substitute x = 1 into (1) and solve for y
2(1)-3y=5
3y = 2 - 5
3y = -3
y = -1
Check the answers in the 2nd equation
3x-(2y-3)/5=4
3(1) - (2(-1)-3)/5 = 4
3 - (-5)/5 = 4
3 - (-1) = 4
4 = 4 Checks...
So our answer is (1, -1)
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calendarEducator since 2010
write12,545 answers
starTop subjects are Math, Science, and Business
We have to solve by elimination the set of equations:
2x - 3y = 5 ...(1)
3x - (2y-3)/5 = 4 ...(2)
3*(1) - 2*(2)
=> 6x - 9y - 6x + 2*(2y-3)/5 = 15 - 8
=> -9y + (2/5)(2y - 3) = 7
=> -45y + 4y - 6 = 35
=> -41y = 41
=> y = -1
(2)
=> 15x - 2y + 3 = 20
=> 15x - 2y = 17 ...(3)
2*(1) - 3*(3)
=> 4x - 6y - 45x + 6y = 10 - 51
=> -41x = -41
=> x = 1
The solution is x = 1 and y = -1
2x - 3y = 5 ---------------(i)
3x - (2y-3)/5 = 4 -------(ii)
From eqn ---( i )
3y = 2x-5
=> y =(2x-5)/3 ---- (iii)
=>2y = (4x-10)/3
=>2y - 3 = (4x - 10)/3 - 3
= (4x -10- 9)/3
= ( 4x -19 )/3
3x - (2y-3)/5 = 4
Now, 3x - (4x - 19 )/15 = 4
=> 45x - 4x +19 = 60
=> 41x= 41
=> x= 1
From eqn ---(iii)
y= (2*1 - 5 )/3
= -3/3 = - 1
(x,y) = (1, -1) -------- solutiom.
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