Question

How do you solve the following system: \[x - 3y = - 1\], \[7x + 15y = 32\]?

Answer 1

Hint: Here in this question, given the system of linear equations. We have to find the unknown values that are \[x\] and \[y\] solving these equations by using the elimination method. In elimination methods either we add or subtract the equations to find the unknown values of\[x\] and \[y\].

Complete step-by-step solution:
Let us consider the equation and we will name it as (1) and (2)
\[x - 3y = - 1\]----------(1)
\[7x + 15y = 32\]-------(2)
Now we have to solve these two equations to find the unknown
Multiply equation (1) with 5, then we get
\[5x - 15y = - 5\]
\[7x + 15y = 32\]
Since the coordinates of \[y\] are same then no need of change the sign here and we simplify to known the unknown value \[x\]
\[
   + 5x - 15y = - 5 \\
  \underline {+ 7x + 15y = 32} \\
  \]
Now we cancel the \[y\] term so we have
\[
   + 5x - 15y = - 5 \\
\underline {+ 7x + 15y = 32} \\
    \\
  12x = 27 \\
 \]
Divide 12 on both sides, then
\[\therefore \,\,\,\,x = \dfrac{{27}}{{12}}\]
We have found the value of \[x\] now we have to find the value of \[y\] . so we will substitute the value \[x\]to any one of the equation (1) or (2) . we will substitute the value of \[x\]to equation (1).
Therefore, we have \[x - 3y = - 1\]
 \[ \Rightarrow \dfrac{{27}}{{12}} - 3y = - 1\]
Isolate y term to the LHS
\[ \Rightarrow 3y = \dfrac{{27}}{{12}} + 1\]
Take 12 as LCM in RHS
\[ \Rightarrow 3y = \dfrac{{27 + 12}}{{12}}\]
\[ \Rightarrow 3y = \dfrac{{39}}{{12}}\]
\[ \Rightarrow 3y = \dfrac{{13}}{4}\]
Divide both side by 3
\[\therefore \,\,y = \dfrac{{13}}{{12}}\]
Hence, we got the unknown values \[x\] and \[y\] that is \[\dfrac{{27}}{{12}}\] and \[\dfrac{{13}}{{12}}\] respectively,
We can check whether these values are correct or not by substituting the unknown values in the given equations and we have to prove L.H.S is equal to R.H.S
Now we will substitute the value of \[x\] and \[y\] in equation (1) so we have
\[\Rightarrow x - 3y = - 1\]
\[ \Rightarrow \dfrac{{27}}{{12}} - 3\left( {\dfrac{{13}}{{12}}} \right) = - 1\]
\[ \Rightarrow \dfrac{{27 - 39}}{{12}} = - 1\]
Multiply both side by 12, we get
\[ \Rightarrow - 12 = - 12\]
Divide both side by -12, then
\[ \Rightarrow \,\,\,1 = 1\]
\[\therefore \,\,\,LHS = RHS\]
Hence the values of the unknown that are \[x\] and \[y\] are the correct values which satisfy the equation.

Note: In this type of question while eliminating the term we must be aware of the sign where we change the sign by the alternate sign. In this we have a chance to verify our answers. In the elimination method we have made the one term have the same coefficient such that it will be easy to solve the equation.