If you mean: 3y^2-7y+3-5y+3-4y^2 then it is simplified to -y^2-12y+6
3y-8+5y = 2y-39 3y+5y-2y = -39+8 6y = -31 y = -31/6
That would depend on the plus or minus value of 3y or maybe an equality is missing none of which have been given.
y=x+5
12x - 9y = 6, 12x + 20y = 64 29y = 58 y = 2 x = 2
5y = 3y + 12 5y - 3y = (3y - 3y) + 12 2y = 12 y = 6
If you mean: 3y^2-7y+3-5y+3-4y^2 then it is simplified to -y^2-12y+6
3y-8+5y = 2y-39 3y+5y-2y = -39+8 6y = -31 y = -31/6
p + q + r = (2x - 9y) + (5y + 6 - 4x) + (3x + 3y - 5) = x - y + 1
That would depend on the plus or minus value of 3y or maybe an equality is missing none of which have been given.
y=x+5
x + 3y = 62x + y = -8Solve by substitution:x + 3y = 6 subtract 3y to both sidesx = 6 - 3y2x + y = -8 substitute 6 - 3y for x2(6 - 3y) + y = -812 - 6y + y = -812 - 5y = -8 subtract 12 to both sides-5y = -20 divide by -5 to both sidesy = 4x = 6 - 3yx = 6 - 3(4) = 6 - 12x = -6Thus, the solution of the system is (-6, 4).
i think its 3y squared
12x - 9y = 6, 12x + 20y = 64 29y = 58 y = 2 x = 2
-4(7x-3y-z)-(-6)(9x-z+2)
3(3y + 2) + 8y 9y + 6 + 8y 17y + 6
5y-7 = 2y+11 (+7 to both sides) 5y = 2y+18 (- 2y from both sides) 3y = 18 (divide both sides by 3 ) y = 6 y=6