SOLUTION: Please help me to solve by substitution or elimination method:
(1). 3x-2y=26
-7x+3y=-49
(2). 4x-5y=14
-12x+15y=-42
(3). -2x+6y=19
10x-30y=-15
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-> SOLUTION: Please help me to solve by substitution or elimination method:
(1). 3x-2y=26
-7x+3y=-49
(2). 4x-5y=14
-12x+15y=-42
(3). -2x+6y=19
10x-30y=-15
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Question 203310: Please help me to solve by substitution or elimination method:
(1). 3x-2y=26
-7x+3y=-49
(2). 4x-5y=14
-12x+15y=-42
(3). -2x+6y=19
10x-30y=-15 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Please help me to solve by substitution or elimination method:
(1).
3x-2y=26
-7x+3y=-49
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Multiply 1st equation by 3; Multiply 2nd equation by 2.
9x - 6y = 78
-14x + 6y = -98
--
Add the results to get rid of the y-terms:
-5x = -20
x = 4
---
Substitute into one of the equations to solve for "y":
3*4 - 2y = 26
-2y = 14
y = -7
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(2).
4x -5y = 14
-12x +15y=-42
---
Multiply thru the 1st by 3:
12x - 15y = 42
---
Add that equation to the 2nd (above) to get:
0 = 0
---
That means the equations are really just one equation.
The solution for the "system" is (x,(4/5)x - (14/5))
=========================================================
(3). -2x+6y=19
10x-30y=-15
--------
Multiply thru the 1st equation by 5 to get:
-10x + 30y = 95
---
Add that to the 2nd equation to get:
0 = 80
That is a contradiction and means there
is no solution. The graphs would be parallel.
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Cheers,
Stan H.
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