SOLUTION: How to solve
1.
2x+3y=13
3x+5y=40
2. 3x+5y=7
4x+6y=12
Algebra.Com
Question 1057488: How to solve
1.
2x+3y=13
3x+5y=40
2. 3x+5y=7
4x+6y=12
Answer by Cromlix(4381) (Show Source): You can put this solution on YOUR website!
Hi there,
2x+3y=13........Eq(1)
3x+5y=40........Eq(2)
Multiply Eq(1) by 3
Multiply Eq(2) by 2
6x + 9y = 39.....Eq(1)
6x +10y = 80.....Eq(2)
Subtract Eq(1) from Eq(20
......y = 41
Substitute y = 41 into Eq(10
2x + 3y = 13
2x + 123 = 13
2x = 13 - 123
2x = -110
x = -55 [x = -55 and y = 41]
..............
3x+5y=7.........Eq(1)
4x+6y=12........Eq(2)
Multiply Eq(1) by 4
Multiply Eq(2) by 3
12x + 20y = 28......Eq(1)
12x + 18y = 36......Eq(2)
Subtract Eq(2) from Eq(1)
.......2y = -8
........y = -4
Substitute y = -4 into Eq(2)
4x + 6y = 12.......Eq(2)
4x + (-24) = 12
4x - 24 = 12
4x = 12 + 24
4x = 36
x = 9 [x = 9 and y = -4]
Hope this helps :-)
RELATED QUESTIONS
Solve by elimination:
1.8x + 11y= 20
5x - 11y= -59
2. 2x + 18y= -9
4x + 18y= -27
(answered by jim_thompson5910)
What is the solution to 1. y = 3x – 8
y = 4 – x
2. x + y = 0
3x + y = -8
(answered by Alan3354)
How would I solve these systems?
1)x+5y=4
3x-7y=-10
2)-5x+3y=6
x-y=4
(answered by elima)
solve each system by any method
1 x-3y=1
3x-5y=-5
2 2x-3y=5... (answered by jim_thompson5910)
Goodevening!!! I have the new problems again and i beg you to answer me more clearly.
(answered by xcentaur)
Solve the system using the substitution method. Show all your steps.
1. x+2y=6... (answered by checkley79)
solve using the addition method.
#1
x-y=6
3x+y=-2
#2
3x-4y=7
x+4y=5
(answered by NinangS)
Use elimination to solve:
3x+5y=-13... (answered by Cromlix)
3x-2y-1=2x-5y-10=4x-3y sOLVE... (answered by Alan3354,Edwin McCravy)