Question

Solve the following pair of equations by substitution method.
\[\begin{array}{l}7x - 15y = 2\\x + 2y = 3\end{array}\]

Answer 1

Hint:
Here we will find the value of \[x\] from the second equation and then we will substitute that value in the first equation. Then we will solve the equation to get the value of \[y\]. Finally, we will put the value of \[y\] in the second equation to get the required answer.

Complete step by step solution:
The two pair of equations given to us is:
\[7x - 15y = 2\]…..\[\left( 1 \right)\]
\[x + 2y = 3\]……\[\left( 2 \right)\]
Now, from equation \[\left( 2 \right)\] we get our \[x\] as,
\[x = 3 - 2y\]…..\[\left( 3 \right)\]
Now we will substitute the value of \[x\] from equation \[\left( 3 \right)\] to equation \[\left( 1 \right)\] and get,
\[7\left( {3 - 2y} \right) - 15y = 2\]
Multiplying the terms using distributive property, we get
\[ \Rightarrow 21 - 14y - 15y = 2\]
Adding the like terms, we get
\[ \Rightarrow 21 - 29y = 2\]
Subtracting 21 from both sides, we get
\[ - 29y = 2 - 21\]
Dividing both sides by \[ - 21\], we get
\[ \Rightarrow y = \dfrac{{ - 19}}{{ - 29}} = \dfrac{{19}}{{29}}\]
Now, we will substitute the above value of \[y\] in equation \[\left( 3 \right)\]. Therefore, we get
\[x = 3 - 2 \times \dfrac{{19}}{{29}}\]
Taking LCM, we get
\[ \Rightarrow x = \dfrac{{3 \times 29 - 2 \times 19}}{{29}}\]
Multiplying the terms, we get
\[ \Rightarrow x = \dfrac{{87 - 38}}{{29}} = \dfrac{{49}}{{29}}\]

So, we get \[x = \dfrac{{49}}{{29}}\] and \[y = \dfrac{{19}}{{29}}\].

Note:
Solving the simultaneous Linear equation can be done by two method Algebraic method and the Graphical method. The substitution method comes under the category of the Algebraic method. The substitution method is used to find the value of the variable by solving two simultaneous Linear equations having two unknown variables. We take the value of one variable from one equation and substitute it in another equation which makes the equation linear with only one unknown variable. We can take the value of \[x\] or \[y\] whichever we want to, they both will give the same result but we always try to take the variable whose value doesn’t contain fraction terms.