many solutions
Divide both sides by 2: 2x = 10 - 2y __ __ ___ 2 2 2 x = 5 - y
4x + 2y = 20 2y = -4x + 20 y = -2x + 10.......but you may also write it as y = 10 - 2x.
(12, 1)
10-2y
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Any pair of numbers where the 'y' is 5 more than the 'x'. There are an infinite number of suitable pairs. 2y-2x=10 2y-2x+2x=10+2x 2y=2x+10 divide both side by 2. y=x+5
2y = -10y = -58 - 2x - 2(-5) = -108 - 2x + 10 = -10-2x = -10 - 182x = 28x = 14Check:2y = 8 - 2x - 2y = -102(-5) = 8 - 2(14) - 2(-5) = -10-10 = 8 - 28 + 10 = -10-10 = -20 + 10 = -10-10 = -10 = -10
many solutions
1 solution
Divide both sides by 2: 2x = 10 - 2y __ __ ___ 2 2 2 x = 5 - y
4x + 2y = 20 2y = -4x + 20 y = -2x + 10.......but you may also write it as y = 10 - 2x.
(12, 1)
One solution 2x+y =5 x+2y=4 multiply 1st eq by 2 rhen subtract: 4x+2y = 10 x + 2y = 4 3x = 6 x = 2 plug x into any of the above two equations and solve y = 1
10-2y
-2x=2y+5 +2x -2y -2y=2x+5 /-2 y=-1/1+2.5
If: 3x+2y = 5x+2y = 14 Then: 3x+2y = 14 and 5x+2y =14 Subtract the 1st equation from the 2nd equation: 2x = 0 Therefore by substitution the solutions are: x = 0 and y = 7