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Four years ago I was $11$ times as old as my dog. Now I am $6$ times as old. How old am I?

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Add, subtract, multiply, and divide decimals: word problems

Full solution

Q. Four years ago I was

11

times as old as my dog. Now I am

6

times as old. How old am I?

Translate Conditions into Equations: Let's denote my current age as $x$ and my dog's current age as $y$ . We are given two conditions: $\newline$ $1$ . Four years ago, I was eleven times as old as my dog. $\newline$ $2$ . Now, I am six times as old as my dog. $\newline$ We can translate these conditions into two equations: $\newline$ Equation $1$ (from condition $1$ ): $x - 4 = 11 \times (y - 4)$ $\newline$ Equation $2$ (from condition $2$ ): $x = 6 \times y$
Solve for Relationship: Now, let's solve Equation $2$ for ' $x$ ' to find the relationship between my age and my dog's age. $\newline$ $x = 6 \times y$ $\newline$ This means that my current age is six times my dog's current age.
Substitute into Equation $1$ : Next, we substitute the expression for $x$ from Equation $2$ into Equation $1$ to find $y$ . $\$6 \times y$ - $4$ = $11$ \times (y - $4$ )\)Now we will solve for $y$ .
Solve for Dog's Age: First, distribute the $11$ on the right side of the equation: $6y - 4 = 11y - 44$ Now, we will move all terms involving 'y' to one side and constants to the other side. $6y - 11y = -44 + 4$ $-5y = -40$
Find My Current Age: Now, we divide both sides by $-5$ to solve for ' $y$ ': $\newline$ $y = -40 / -5$ $\newline$ $y = 8$ $\newline$ So, my dog is currently $8$ years old.
Find My Current Age: Now, we divide both sides by $-5$ to solve for $y$ : $y = \frac{-40}{-5}$ $y = 8$ So, my dog is currently $8$ years old.Now that we know $y$ , we can use Equation $2$ to find $x$ , which is my current age. $x = 6 \times y$ $x = 6 \times 8$ $x = 48$ So, I am currently $48$ years old.

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