Question

1. taylor has a $30.00 gift card that she can spend at the store. she has already bought a $5.00 picture frame, and she wants to buy $3.00 journals with the leftover money on the card. how many journals can she buy with her gift card? inequality: 1.50 + 3.00j ≤ 15 variable: j = journals solution: 4. the length of a rectangle is eight more than twice its width. if the length is less than 34, what is the width? inequality: solution: variable:

Explanation:

Step1: Set up the inequality for the journal - buying problem

Let $J$ be the number of journals. The cost of each journal is $\$3.00$. Taylor has a $\$30.00$ gift - card and has already spent $\$7.50$. So the amount of money she can spend on journals is $30 - 7.50=22.50$. The cost of $J$ journals is $3J$, and the inequality representing the situation is $3J\leqslant22.50$.

Step2: Solve the inequality for $J$

Divide both sides of the inequality $3J\leqslant22.50$ by $3$. We get $J\leqslant\frac{22.50}{3}=7.5$. Since the number of journals must be a non - negative integer, the maximum number of journals she can buy is $7$.

Step3: Set up the equation for the rectangle problem

Let $w$ be the width of the rectangle. The length $l$ of the rectangle is given by $l = 2w+8$. We are also given that $l\lt34$. Substitute $l = 2w + 8$ into the inequality $l\lt34$. We get $2w+8\lt34$.

Step4: Solve the inequality for $w$

Subtract $8$ from both sides of the inequality $2w+8\lt34$: $2w\lt34 - 8=26$. Then divide both sides by $2$: $w\lt13$.

Answer:

For the journal problem, the maximum number of journals Taylor can buy is $7$. For the rectangle problem, the width $w$ of the rectangle satisfies $w\lt13$.