Question

How do you solve the system of equations $3x - 2y = 5$ and $x = y + 2?$

Answer 1

Hint: This problem deals with solving the system of linear equations in $x$ and $y$. This is done by adding or subtracting a multiple of one equation to the other equation, in such a way that either the $x$-terms or the $y$-terms cancel out. Then solve for $x$(or $y$, whichever left) and substitute back to get the other coordinate.

Complete step-by-step solution:
Given a pair of linear equations in $x$ and $y$, which are given by $3x - 2y = 5$ and $x = y + 2$,
Now we have to solve these two equations in order to get the values of $x$ and $y$.
Consider the given two equations and rearrange them to solve accordingly.
Consider the first equation, as given below:
$ \Rightarrow \;3x - 2y = 5$
Now consider the second equation, as given below:
$ \Rightarrow x = y + 2$
Rearranging the above equation, so as to solve this equation together with the first equation, as:
$ \Rightarrow x - y = 2$
Now multiply the above equation with 3, so as match the coefficient of $x$, so that we can solve both the equations in a simpler way, as shown below:
$ \Rightarrow 3\left( {x - y = 2} \right)$
$ \Rightarrow 3x - 3y = 6$
Now subtracting the first and second equation, as shown below:
$ \Rightarrow \;3x - 2y = 5$
$ \Rightarrow 3x - 3y = 6$
$ \Rightarrow - 2y - \left( { - 3y} \right) = 5 - 6$
On simplification of the above two equations, as given below:
$ \Rightarrow - 2y + 3y = - 1$
$\therefore y = - 1$
Substituting the value of $y = - 1$, in the first equation, as given below:
$ \Rightarrow \;3x - 2y = 5$
$ \Rightarrow \;3x - 2\left( { - 1} \right) = 5$
$ \Rightarrow \;3x + 2 = 5$
Grouping the constants to the other side of the equation, gives:
$ \Rightarrow \;3x = 3$
$\therefore x = 1$
Hence the values of $x = 1$ and $y = - 1$.

The values of $x$ and $y$ are 1 and -1 respectively.

Note: Here a system of equations is when we have two or more linear equations working together. Here this problem can be done in another way but with a slight change with the method solved here. Here instead of multiplying the second equation with 3, we can multiply the second equation with -2 and solve both the equations, finally ending up the same solution.