Question

How do you solve the following system: \[x + 3y = - 2\], \[5x + 2y = 5\]?

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)
\[x + 3y = - 2\]----------(1)
\[5x + 2y = 5\]----------(2)
Now we have to solve these two equations to find the unknown
Multiply (1) by 5, then we get
\[5x + 15y = - 10\]
\[5x + 2y = 5\]
Since the coordinates of \[x\] are same and we change the sign by the alternate sign and we simplify to known the unknown value \[y\]
\[
   + 5x + 15y = - 10 \\
\underline { \mathop + \limits_{( - )} 5x\mathop + \limits_{( - )} 2y = \mathop + \limits_{( - )} 5} \\
  \]
Now we cancel the \[x\] term so we have
\[
  + 5x + 15y = - 10 \\
\underline { \mathop + \limits_{( - )} 5x\mathop + \limits_{( - )} 2y = \mathop + \limits_{( - )} 5} \\
    \\
  13y = - 15 \\
 \]
Divide both side by 13, then
\[\therefore \,\,\,\,y = - \dfrac{{15}}{{13}}\]
We have found the value of \[y\] now we have to find the value of \[x\]. so we will substitute the value \[y\]to any one of the equation (1) or (2) . we will substitute the value of \[y\] to equation (1).
Therefore, we have \[x + 3y = - 2\]
 \[ \Rightarrow x + 3\left( { - \dfrac{{15}}{{13}}} \right) = - 2\]
\[ \Rightarrow \dfrac{{13x - 45}}{{13}} = - 2\]
Multiply both side by 13
\[ \Rightarrow 13x - 45 = - 26\]
Add 45 on both side, then
\[ \Rightarrow 13x - 45 + 45 = - 26 + 45\]
\[ \Rightarrow 13x = 19\]
Divide both side by 13, then
\[\therefore \,\,\,\,x = \dfrac{{19}}{{13}}\]
Hence we got the unknown values \[x\] and \[y\] that is \[ - \dfrac{{15}}{{13}}\] and \[\dfrac{{19}}{{13}}\] 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
\[x + 3y = - 2\]
\[ \Rightarrow \dfrac{{19}}{{13}} + 3\left( { - \dfrac{{15}}{{13}}} \right) = - 2\]
\[ \Rightarrow \dfrac{{19 - 45}}{{13}} = - 2\]
\[ \Rightarrow \dfrac{{ - 26}}{{13}} = - 2\]
Multiply 13 on both side, then
\[ \Rightarrow - 26 = - 26\]
Divide both side by -26
\[ \Rightarrow 1 = 1\]
\[\therefore \,\,\,LHS = RHS\]
Hence the values of the unknown that is \[x\] 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.