Question

How do you solve the system of linear equations $ - 5x - 3y = 11{\text{ and }}2x - 4y = 15$?

Answer 1

Hint: This type of problem is known as a system pair of linear equations in two variables. The general form of the linear equation is $ax + by + c = 0$. The pair of linear equations in two variables are in the same two variables ${\text{x and y}}$. The general form for a pair of linear equations in two variable ${\text{x and y}}$ is ${{\text{a}}_{\text{1}}}{\text{x + }}{{\text{b}}_{\text{1}}}{\text{y + }}{{\text{c}}_{\text{1}}}{\text{ = 0 and }}{{\text{a}}_{\text{2}}}{\text{x + }}{{\text{b}}_{\text{2}}}{\text{y + }}{{\text{c}}_{\text{2}}}{\text{ = 0}}$, where ${{\text{a}}_{\text{1}}}{\text{,}}{{\text{b}}_{\text{1}}}{\text{,}}{{\text{c}}_{\text{1}}}{\text{,}}{{\text{a}}_{\text{2}}}{\text{,}}{{\text{b}}_{\text{2}}}{\text{,}}{{\text{c}}_{\text{2}}}{\text{ = 0}}$ are real numbers. Here we solve by substitution method. From two equations we separate the one variable and then with that we substitute in another equation we will find the required values and also by complete step by step explanation.

Complete step-by-step solution:
Let us consider the equations: $ - 5x - 3y = 11 - - - - (1)$ and $2x - 4y = 15 - - - - (2)$
Now separate variable $x$ in equation (1)
$ \Rightarrow - 5x - 3y = 11$
On rewriting the term and we get,
$ \Rightarrow - 5x = 11 + 3y$
Let us divide 5 on both sides we get,
$ \Rightarrow x = \dfrac{{11 + 3y}}{{ - 5}} - - - - (3)$
Now substitute equation (3) in equation (2)
$ \Rightarrow 2x - 4y = 15$
$ \Rightarrow 2\left( {\dfrac{{11 + 3y}}{{ - 5}}} \right) - 4y = 15$
On simplify the terms and we get,
$ \Rightarrow \dfrac{{22 + 6y}}{{ - 5}} - 4y = 15$
Now taking Least Count Multiple (LCM) on Left Hand Side (LHS), we get
$ \Rightarrow \dfrac{{22 + 6y + 20y}}{{ - 5}} = 15$
Taking cross multiplication on we get
$ \Rightarrow 22 + 6y + 20y = 15 \times - 5$
Let us multiply we get,
$ \Rightarrow 22 + 26y = - 75$
On rewriting we get,
$ \Rightarrow 26y = - 75 - 22$
On adding we get,
$ \Rightarrow 26y = - 97$
Let us divide the terms
$ \Rightarrow y = \dfrac{{ - 97}}{{26}}$
Now substitute value of $y$ in equation (2)
\[ \Rightarrow 2x - 4y = 15\]
\[ \Rightarrow 2x - 4\left( {\dfrac{{ - 97}}{{26}}} \right) = 15\]
Now taking Least Count Multiple (LCM) on Left Hand Side (LHS), we get
\[ \Rightarrow 26x + 194 = 195\]
\[ \Rightarrow \dfrac{{26x + 194}}{{13}} = 15\]
Let us cross multiply we get,
\[ \Rightarrow 26x + 194 = 15 \times 13\]
On multiply we get
\[ \Rightarrow 26x + 194 = 195\]
On subtracting we get,
$ \Rightarrow 26x = 195 - 194$
$ \Rightarrow 26x = 1$
$ \Rightarrow x = \dfrac{1}{{26}}$
Thus we obtain the value of $x$ and $y$ $ \Rightarrow x = \dfrac{1}{{26}}$

Therefore the value of x is $\dfrac{1}{{26}}$ and y is $\dfrac{{ - 97}}{{26}}$.

Note: The pair of linear equations in two variables has four methods to solve. They are graphical method, elimination method, substitution method and cross multiplication method.
We can identify the equations are consistent or inconsistent and we also find the equations of straight lines in graphically they are parallel, intersecting or coincident.
If the coefficients of equation $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$are intersecting graphically they have unique solutions.
If the coefficients of equation $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ are coincident graphically they have infinitely many solutions.
If the coefficients of equations $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ are parallel lines graphically they have no solution.