Presumably, x and y have to be positive integers.
If 5x + 7y = 89 then we need to find a multiple of 7 that when subtracted from 84 leaves a number ending in 0 or 5, thus allowing this remainder number to be divisible by 5.
7 x 1 = 7 : 89 - 7 = 82
7 x 2 = 14 : 89 - 14 = 75........which meets the above requirement
Then, 5x = 75 : x = 15.
After this first multiple of 7 has been determined, further multiples of 7 must be formed from an increase of 5, 10, 15 etc to enable the remainder to be divisible by 5. In a similar way, further multiples of 5 must REDUCE by multiples of 7. This means that there are only two further values for x, namely x = 15 - 7 = 8 and x = 15 - 14 = 1.
So, 7 x (2 + 5) = 7 x 7 = 49 : 89 - 49 = 40 then 5x = 40, : x = 8
And, 7 x (2 + 10) = 7 x 12 = 84 : 89 - 84 = 5 then 5x = 5 : x = 1.
The three possible solutions are (x = 15, y = 2), (x = 8, y = 7) and (x = 1, y = 12).
As mentioned above, note that the values for x decrease by 7 and the values for y increase by 5 for successive solutions.
if: 2x + 6 = -6 then: 2x = -12 x = -6 if: 5x + 7y = 1 then: 5(-6) + 7y = 1 7y = 31 y = 31/7
1 + 7y = 5x - 2 7y = 5x - 3 - 5x = - 7y - 3 x = 7/5y + 3/5 ---------------------
(-4, -1)
2x + 3y = 17 3x + 4y = 24 (2x + 3x) + (3y + 4y) = (17 + 24) 5x + 7y = 41
5y explanaition 5x-5x=0, 0-2y=-2y, -2y+7y=5y
Combine like terms: -2x - 4y + 5x - 7y = 3x - 11y -2x + 5x = 3x -4y + -7y = -11y
if: 2x + 6 = -6 then: 2x = -12 x = -6 if: 5x + 7y = 1 then: 5(-6) + 7y = 1 7y = 31 y = 31/7
1 + 7y = 5x - 2 7y = 5x - 3 - 5x = - 7y - 3 x = 7/5y + 3/5 ---------------------
(-4, -1)
-5x+7y-6z
2x + 3y = 17 3x + 4y = 24 (2x + 3x) + (3y + 4y) = (17 + 24) 5x + 7y = 41
5y explanaition 5x-5x=0, 0-2y=-2y, -2y+7y=5y
9x + 3y + 5z - 5x + 4z - 7y = 4x - 4y + 9z
(4, -7)
It works out that they intersect at: (4, -7)
Yes it is.
5x - 7y = 5 (subtract both sides by 5x) -7y = -5x + 5 (divide both sides by -7) y = (5x - 5)/7 If you want to solve for y = 0, then x = 1.