What is the answer to 7(3x+1)+8(2x-3)=4(3x-1)-7(x-4) ?

The answer to 7(3x+1)+8(2x-3)=4(3x-1)-7(x-4) is x= 41/32

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7(3x+1)+8(2x-3)=4(3x-1)-7(x-4)

7 (3 x + 1) + 8 (2 x - 3) = 4 (3 x - 1) - 7 (x - 4)

x = \frac{41}{32}

Solution steps

7 (3 x + 1) + 8 (2 x - 3) = 4 (3 x - 1) - 7 (x - 4)

Expand

7 (3 x + 1) + 8 (2 x - 3) : 37 x - 17

Expand

4 (3 x - 1) - 7 (x - 4) : 5 x + 24

37 x - 17 = 5 x + 24

Move

17

to the right side

37 x = 5 x + 41

Move

5 x

to the left side

32 x = 41

Divide both sides by

32

x = \frac{41}{32}

e^{- l n (1 - y)}

\frac{6}{7} x - \frac{1}{4} \leq \frac{11}{14} x - \frac{4}{7}

z^{2} - 14 z + 24

7 (v + 5) = 105

- 4 x \geq 28

What is the answer to 7(3x+1)+8(2x-3)=4(3x-1)-7(x-4) ?
The answer to 7(3x+1)+8(2x-3)=4(3x-1)-7(x-4) is x= 41/32
Solve for x: 7(3x+1)+8(2x-3)=4(3x-1)-7(x-4)
The solution is x= 41/32