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Question
name: block 4 algebra i sped teacher resource room math coach - ms. almonte 9. solve this inequality: $\frac{x - 4}{3}geq\frac{x + 3}{2}$
Step1: Cross - multiply
Multiply both sides by 6 to clear the fractions. $6\times\frac{x - 4}{3}\leq6\times\frac{x + 3}{2}$.
$2(x - 4)\leq3(x + 3)$
Step2: Expand brackets
Use the distributive property. $2x-8\leq3x + 9$
Step3: Isolate x terms
Subtract 2x from both sides. $2x-2x - 8\leq3x-2x + 9$
$-8\leq x + 9$
Step4: Solve for x
Subtract 9 from both sides. $-8-9\leq x+9 - 9$
$x\geq - 17$
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$x\geq - 17$