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
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2y - 2x = 10 is:
2y = 10 + 2x
2y/2 = (10 + 2x)/2
y = x + 5
-2x = -2y + 10
-2x/-2 = (-2y + 10)/-2
x = y - 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
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
y = 0 and x = 8.
Simultaneous suggests at least two equations.
16x - 2y = 74 (Eq 1) 2x - 2y = 4 (Eq 2) from Eq 2, 2y = 2x - 4 Substitute in Eq 1 16x - (2x - 4) =74 ie 16x - 2x + 4 = 74 ie 14x = 70 ie x = 5 2y = 2x - 4 = 10 - 4 = 6 y =3 Check 16 x 5 - 2 x 3 = 80 - 6 = 84 QED