Question

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." Represent this situation algebraically.

Answer 1

Hint: A linear system of two equations with two variables is any system that can be written in the form.
\[
  ax + by = p \\
  cx + dy = q \\
 \]
where any of the constants can be zero with the exception that each equation must have at least one variable in it.
we use the Linear equation with two variables to solve the question.

Complete step by step solution:
A way to solve a linear system algebraically is to use the substitution method. The substitution method functions by substituting the one y-value with the other.
Let Aftab’s age be = x
His daughter’s age be =y
Then , seven years ago,
Aftab's age =x−7
His daughter's age =y−7
According to the question, Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then.
\[ \Rightarrow x - 7 = 7\left( {y - 7} \right)\]
\[ \Rightarrow x - 7 = 7y - 49\]
\[ \Rightarrow x - 7y = - 49 + 7\]
\[ \Rightarrow x - 7y = - 42\]
\[ \Rightarrow 7y - x = 42\]
Equation 1
After three years,
Aftab's age =$x+3$
His daughter's age =$y+3$
According to the question, Aftab tells his daughter “three years from now, I shall be three times as old as you will be.”
\[ \Rightarrow x + 3 = 3\left( {y + 3} \right)\]
\[ \Rightarrow x + 3 = 3y + 9\]
\[ \Rightarrow x - 3y = 9 - 3\]
\[ \Rightarrow x - 3y = 6\]
Equation 2
We can substitute $x$ in the second equation with the first equation since $y = y$.
Using equation 2 find the value of $x$ in terms of $y$
\[ \Rightarrow x = 3y + 6\]
In this equation, we have isolated the variable $x$.
\[ \Rightarrow 7y - x = 42\]
Now, we substitute the $x$ in equation 1 with the expression for the second equation.
\[ \Rightarrow 7y - \left( {3y + 6} \right) = 42\]
\[ \Rightarrow 7y - 3y - 6 = 42\]
\[ \Rightarrow 4y = 42 + 6\]
\[ \Rightarrow 4y = 48\]
\[ \Rightarrow y = \dfrac{{48}}{4} = 12\]
To determine the $x$-value, we may proceed by inserting our $y$-value in any of the equations. We select the second equation:
\[ \Rightarrow x = 3y + 6\]
\[ \Rightarrow x = 3\left( {12} \right) + 6\]
\[ \Rightarrow x = 36 + 6 = 42\]

Note:
You can use the substitution method even if both equations of the linear system are in standard form. Just begin by solving one of the equations for one of its variables.
You must check your solutions, in every linear equation. To do this "checking", you need only plug your answer into the original equation, and make sure that you end up with a true statement.
Substituting the value of $x=42$ and $y=12$ in equation (1)
\[ \Rightarrow x - 3y = 6\]
$ \Rightarrow 42 - 3(12) = 42 - 36 = 6$
So, therefore my L.H.S. = R.H.S. . Hence our solution is absolutely correct.