What is the value of $x$ if $-\frac23(x-5) = \frac32(x+1)$?

x = (-29)/5

Step-by-step explanation:

Solve for x:

(2 (x - 5))/3 = (3 (x + 1))/2

Multiply both sides by 6:

(6×2 (x - 5))/3 = (6×3 (x + 1))/2

6/3 = (3×2)/3 = 2:

2×2 (x - 5) = (6×3 (x + 1))/2

6/2 = (2×3)/2 = 3:

2×2 (x - 5) = 3×3 (x + 1)

2×2 = 4:

4 (x - 5) = 3×3 (x + 1)

3×3 = 9:

4 (x - 5) = 9 (x + 1)

Expand out terms of the left hand side:

4 x - 20 = 9 (x + 1)

Expand out terms of the right hand side:

4 x - 20 = 9 x + 9

Subtract 9 x from both sides:

(4 x - 9 x) - 20 = (9 x - 9 x) + 9

4 x - 9 x = -5 x:

-5 x - 20 = (9 x - 9 x) + 9

9 x - 9 x = 0:

-5 x - 20 = 9

Add 20 to both sides:

(20 - 20) - 5 x = 20 + 9

20 - 20 = 0:

-5 x = 9 + 20

9 + 20 = 29:

-5 x = 29

Divide both sides of -5 x = 29 by -5:

(-5 x)/(-5) = 29/(-5)

(-5)/(-5) = 1:

x = 29/(-5)

Multiply numerator and denominator of 29/(-5) by -1:

x = (-29)/5

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3