What is the answer to (4z+1)/3+4=(z+3)/2+2(z+2) ?

The answer to (4z+1)/3+4=(z+3)/2+2(z+2) is z=-1

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(4z+1)/3+4=(z+3)/2+2(z+2)

\frac{4 z + 1}{3} + 4 = \frac{z + 3}{2} + 2 (z + 2)

z = - 1

Solution steps

\frac{4 z + 1}{3} + 4 = \frac{z + 3}{2} + 2 (z + 2)

Multiply by LCM

2 (4 z + 1) + 24 = 3 (z + 3) + 12 (z + 2)

Expand

2 (4 z + 1) + 24 : 8 z + 26

Expand

3 (z + 3) + 12 (z + 2) : 15 z + 33

8 z + 26 = 15 z + 33

Move

26

to the right side

8 z = 15 z + 7

Move

15 z

to the left side

- 7 z = 7

Divide both sides by

- 7

z = - 1

x, [(x - 3)] \frac{2}{16} + [(y - 1)] \frac{2}{25} = 1

x, 2 x - 2 y = - 6

x, 2 x - 2 y = - 2

x, 3 x - y = 5 y

\frac{3}{4} = 6

What is the answer to (4z+1)/3+4=(z+3)/2+2(z+2) ?
The answer to (4z+1)/3+4=(z+3)/2+2(z+2) is z=-1
Solve for z: (4z+1)/3+4=(z+3)/2+2(z+2)
The solution is z=-1