How to solve 4(z+3)=7z+40

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 4(z+3)=7z+40 | Equatix

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Question

$$4(z+3)=7z+40$$

Solve for z

$z = -\frac{28}{3} = -9\frac{1}{3} \approx -9.333333333$

Steps for Solving Linear Equation

Use the distributive property to multiply $4$ by $z+3$.

$$4z+12=7z+40$$

Subtract $7z$ from both sides.

$$4z+12-7z=40$$

Combine $4z$ and $-7z$ to get $-3z$.

$$-3z+12=40$$

Subtract $12$ from both sides.

$$-3z=40-12$$

Subtract $12$ from $40$ to get $28$.

$$-3z=28$$

Divide both sides by $-3$.

$$z=\frac{28}{-3}$$

Fraction $\frac{28}{-3}$ can be rewritten as $-\frac{28}{3}$ by extracting the negative sign.

$$z=-\frac{28}{3}$$

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Solve for z

Steps for Solving Linear Equation