How do you solve 4y plus x equals 8
6x = 32 - 4y and 4y - 8 = 6x so 32 - 4y = 4y - 8 ie 40 = 8y so y = 5 and x = 2
x = 4 and y = 0
By definition: 4y-2 = 6y+8 4y-6y = 8+2 -2y = 10 y = -5 and by substituting this value into the equation x = -22
3 x 3 + 4 x 8 = 9 + 32 ...
How do you solve 4y plus x equals 8
6x = 32 - 4y and 4y - 8 = 6x so 32 - 4y = 4y - 8 ie 40 = 8y so y = 5 and x = 2
x = 4 and y = 0
By definition: 4y-2 = 6y+8 4y-6y = 8+2 -2y = 10 y = -5 and by substituting this value into the equation x = -22
(8 + 4) x 5...
3 x 3 + 4 x 8 = 9 + 32 ...
x=4y+1 x=4y-1 No,they have different solutions.
No.
If x - 4y = 2 and x + 4y = 2 then the only solution is when y = 0 and thus x = 2.
Multiply the first equation by -3
x = -4y so 3x = -12y Substituting in the second equation, -12y + 2y = 20 or -10y = 20 ie y = -2 And then x = -4y implies that x = 8 Solution: x = 8, y = -2
2x - 4y = 8 2x = 4y + 8 x = 2y + 4