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Leo had \[tex]$91, which is 7 times as much money as Alison had. How much money did Alison have?

Select the correct solution method below, where \( x \) represents Alison's money.

A. \( \frac{x}{7} = 91 \). Multiply both sides by 7. Alison had \$[/tex]637.

B. [tex]\( x - 7 = 91 \)[/tex]. Add 7 to both sides. Alison had \[tex]$98.

C. \( x + 7 = 91 \). Subtract 7 from both sides. Alison had \$[/tex]84.

D. [tex]\( 7x = 91 \)[/tex]. Divide both sides by 7. Alison had \$13.

Sagot :

GinnyAnsweridn
Let's solve this step-by-step.

We know that Leo had \[tex]$91, which is 7 times as much money as Alison had. We want to find the amount of money Alison had, denoted by \(x\). The equation representing this situation is: \[ 7x = 91 \] To solve for \(x\), we need to isolate \(x\) on one side of the equation. We do this by dividing both sides of the equation by 7: \[ x = \frac{91}{7} \] When we perform this division, we get: \[ x = 13 \] Therefore, Alison had \$[/tex]13.

Looking at the solution methods provided:

A. [tex]\(\frac{x}{7} = 91\)[/tex]. Multiply both sides by 7. Alison had \[tex]$637. (Incorrect). B. \(x - 7 = 91\). Add 7 to both sides. Alison had \$[/tex]98. (Incorrect).
C. [tex]\(x + 7 = 91\)[/tex]. Subtract 7 from both sides. Alison had \[tex]$84. (Incorrect). D. \(7x = 91\). Divide both sides by 7. Alison had \$[/tex]13. (Correct).

So, the correct solution method is D.
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