How to solve 7 > 8x-5

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 7 > 8x-5 | Equatix

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$$7 > 8x-5$$

$x<\frac{3}{2}$

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$8x-5<7$$

Add $5$ to both sides.

$$8x<7+5$$

Add $7$ and $5$ to get $12$.

$$8x<12$$

Divide both sides by $8$. Since $8$ is positive, the inequality direction remains the same.

$$x<\frac{12}{8}$$

Reduce the fraction $\frac{12}{8}$ to lowest terms by extracting and canceling out $4$.

$$x<\frac{3}{2}$$