Question

1. company a hires employees for $11.50 an hour. company b hires employees for $10 an hour but offers an $800 signing bonus for all new employees. a. write an inequality to represent when it is better to work at company b than company a. use x to represent the hours worked. b. solve the inequality. interpret your solution in the context of this problem.

Explanation:

Step1: Set up the cost - functions

Let $x$ be the number of hours worked. The total earnings at Company A is $E_A = 800+11.5x$, and the total earnings at Company B is $E_B = 10x$.

Step2: Write the inequality

We want to find when it is better to work at Company B than Company A, so we set up the inequality $E_B>E_A$. Substituting the expressions for $E_A$ and $E_B$, we get $10x>800 + 11.5x$.

Step3: Solve the inequality

Subtract $11.5x$ from both sides: $10x-11.5x>800+11.5x - 11.5x$. This simplifies to $- 1.5x>800$. Divide both sides by $-1.5$. When dividing an inequality by a negative number, the direction of the inequality sign changes. So $x<-\frac{800}{1.5}=-\frac{1600}{3}\approx - 533.33$. But the number of hours worked $x\geq0$. Let's set up the correct inequality $E_A800 + 11.5x$ wrong. The correct one is $10x<800 + 11.5x$. Subtract $10x$ from both sides: $0<800+11.5x - 10x$, which simplifies to $0<800 + 1.5x$. Subtract 800 from both sides: $-800<1.5x$. Divide both sides by 1.5: $x>\frac{-800}{1.5}=-\frac{1600}{3}$. Since $x\geq0$, the correct inequality is $10x<800 + 11.5x$. Solving for $x$ gives $x > \frac{800}{1.5}=\frac{1600}{3}\approx533.33$.

Answer:

a. The inequality is $10x<800 + 11.5x$.
b. Solving the inequality $10x<800 + 11.5x$:

• Subtract $10x$ from both sides: $0<800+1.5x$.
• Subtract 800 from both sides: $-800 < 1.5x$.
• Divide both sides by 1.5: $x>\frac{800}{1.5}=\frac{1600}{3}\approx533.33$.
• Interpretation: It is better to work at Company B than Company A when the number of hours worked $x$ is greater than $\frac{1600}{3}\approx533.33$ hours.