Question

How do you write the ordered pair that is the solution to the following system of equations: $5x - 2y = 3$ and $3x + 4y = 7$?

Answer 1

Hint: Here, we are required to find the ordered pair that is the solution of the given system of linear equations in two variables. Thus, we will use elimination method (or any other method) to find the values of $x$ and $y$ by eliminating the equations and substituting their values and hence, we will be able to find the required values of $x$ and $y$ which will give us the required ordered pair.

Complete step by step solution:
The given two equations are:
$5x - 2y = 3$………..$ \times 2$
$3x + 4y = 7$
We will solve these equations using the Elimination method by multiplying the first equation by 2 such that the coefficients of variable $y$ in both the equations become the same.
Hence, adding the equations, we get,
$2\left( {5x - 2y} \right) + \left( {3x + 4y} \right) = 2\left( 3 \right) + 7$
$ \Rightarrow 10x - 4y + 3x + 4y = 6 + 7$
Hence, adding the like terms together, we get,
$ \Rightarrow 13x = 13$
Dividing both sides by 13,
\[\]
Now, substituting this value in the equation, $5x - 2y = 3$, we get,
$5\left( 1 \right) - 2y = 3$
$ \Rightarrow 5 - 2y = 3$
Subtracting 3 from both sides and adding $2y$on both sides, we get,
$ \Rightarrow 5 - 2y - 3 + 2y = 3 - 3 + 2y$
$ \Rightarrow 5 - 3 = 2y$
Thus, we get,
$ \Rightarrow 2 = 2y$
Dividing both sides by 2,
$ \Rightarrow y = 1$
Therefore, for the given system of equations $5x - 2y = 3$ and $3x + 4y = 7$, we get $x = 1$and $y = 1$
Hence, the ordered pair $\left( {x,y} \right) = \left( {1,1} \right)$

Hence, the ordered pair that is the solution to the following system of equations: $5x - 2y = 3$ and $3x + 4y = 7$is $\left( {1,1} \right)$
Thus, this is the required answer.

Note:
By substituting the values of $x$and $y$ in the given equations we can check whether our answer is correct or not.
Hence, substituting the value $\left( {x,y} \right) = \left( {1,1} \right)$ in the first equation,
$5x - 2y = 3$
LHS$ = 5\left( 1 \right) - 2\left( 1 \right) = 5 - 2 = 3 = $RHS
Since,
LHS$ = $RHS
Hence, these values satisfy the first equation.
Similarly, substituting these values in the LHS and RHS of the second equation, we would again find them equal. Hence, this shows that our answer is correct and there is not a single step which we have attempted wrong.