Linda received her salary on Friday. She spent one-half of her salary on rent and then spent $268.00 on her electric and water bill. Next she spent one-half of what was left on her phone bill. After spending $52.00 on gas for her car, Linda had $82.00 left. How much did Linda spend on rent? A. $260.00
B. 536.00
C. 618.00
D. 1,072.00

B. $536.00

Step-by-step explanation:

For this problem we need to evaluate top down to build our equation, and then solve bottom up to find how much she spent on rent.

Let Z represent Linda's salary

Let Y represent Linda's Water & Electric (given $268)

Let X represent Linda's Phone Bill

Let W represent Linda's gas bill (given $52)

We are given that after all Linda's expenses, she has $82 left.

So let's find Z/2.

Z/2 = Y + k

Where k is the remainder after Linda pays Y. Hence, k will be equivalent to X + W + 82.

So we can write:

Z/2 = Y + X + W + 82

Using our knowns, we can write in:

Z/2 = 268 + X + 52 + 82

We know that X is (Z/2 - Y)/2 from the sentence, "Next, she spent one-half of what she had left on her phone bill." So we can say X = k/2.

X = (X + W +82)/2

X = X/2 + 52/2 + 82/2

X = X/2 + 26 + 41

X - X/2 = X/2 - X/2 + 67

X/2 = 67

X = 134

So, now we can finish our equation to find Z/2.

Z/2 = 268 + X + 52 + 82

Z/2 = 268 + 134 + 52 + 82

Z/2 = 536

Hence, we have found that Linda spent half of her salary on rent, which was $536.00.

Cheers.

Step-by-step explanation:

To solve this, we need to determine how much Linda's salary was. Then, we can divide the number by 2 to get the answer, as she spent 1/2 of her salary on rent.

The first step is to add up the remaining 1/2 and write an equation.

$268 + 1/2(s - 268) + $52 + $82 = s

Now, we combine like terms.

$402 + 1/2(s - 268) = s

$402 + 1/2s - 1/2(268) = s

$402 + 1/2s - $134 = s

$268 + 1/2s = s

1/2s = s - $268

1/2s = $268

s = $536

Now, we can check our answer by the equation

$402 + 1/2(536) - 1/2(268) = 536

$402 + 268 - 134 = 536

536 = 536 ✅

sites:

answer: i wpould go with c

step-by-step explanation: