Your question cannot be answered. There not even sufficient operations and and equal sign witch the equation equals to.
I don't know whether it's y^2-3y+2+3y-4y-5 or 2y-3y+2+3y-4y-5, so I'll do both.y^2-3y+2+3y-4y-5y^2+(-3y+3y-4y)+(2-5); group them togethery^2+(-4y)+(-3)y2-4y-3OR2y-3y+2+3y-4y-5(2y-3y+3y-4y)+(2-5); group them together(-2y)+(-3)-2y-3
If y + 3y + 10 = 70, then 4y + 10 = 70, and 4y = 60. We then find that y = 60/4 = 15.
10
X + 3y - 2y + 3y = X + (3-2+3)y = X + 4y
1
y = 3.6 (18/5)
2x + 3y = 17 3x + 4y = 24 (2x + 3x) + (3y + 4y) = (17 + 24) 5x + 7y = 41
4y-3y = 1
4y + 7 = 3y + 9-3y-7 -3y-7y = 24(2) + 7 = 3(2) + 98 + 7 = 6 + 915 = 15
I don't know whether it's y^2-3y+2+3y-4y-5 or 2y-3y+2+3y-4y-5, so I'll do both.y^2-3y+2+3y-4y-5y^2+(-3y+3y-4y)+(2-5); group them togethery^2+(-4y)+(-3)y2-4y-3OR2y-3y+2+3y-4y-5(2y-3y+3y-4y)+(2-5); group them together(-2y)+(-3)-2y-3
10x - 3y + 2x - 4y =(10x + 2x) + (-3y - 4y) =12x - 7y
To determine if (3x - 3y = 21) and (5x + 4y = 24) have a solution, we can solve the system of equations. Rearranging the first equation gives (x - y = 7) or (x = y + 7). Substituting this into the second equation allows us to find values for (x) and (y). Solving the resulting equation will reveal whether there are specific values that satisfy both equations simultaneously.
41
4y + 7 < 3y + 9 4y - 3y < 9 - 7 y < 2
-7y + 4y = -3y
4y + 4 = 3y - 5 y = -9
OK, so-3y + 15 - y = 39-3y + 15 - 15 - y = 39 - 15-3y -y = 24-4y = 24-y = 6y= -6.Hope this helps!