Question

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

Answer 1

Hint: Where we are given a pair of linear equations and we have to find the value of x and y by using the given equation. We can solve the equation by the method of elimination or by using the method of substitution for the method of substitution. First we will find the value of one variable in the form of another for example we will find the value of x in terms of y then substitute that value in another equation. Then we will solve the equation and find the value of that variable. After that, substitute the value of that variable in another equation and find the value of the remaining one variable.

Complete step-by-step answer:
Step1: We are given a pair of linear equations $ - 5x - 3y = 11$ and $2x - 4y = 15$ by applying the method of substitution we will find the value of both variables. We will solve the second equation for $x$:
$ \Rightarrow 2x - 4y = 15$
Adding $4y$ from both sides:
$ \Rightarrow 2x + 3y - 4y = 15 + 4y$
On proper rearrangement we will get:
$ \Rightarrow 2x = 15 + 4y$
Now we will divide both the sides by 2 then we get:
$ \Rightarrow x = \dfrac{{15 + 4y}}{2}$
Step2: Now we will substitute the value of x in the first equation and solve for $y$:
$ \Rightarrow - 5\left( {\dfrac{{15 + 4y}}{2}} \right) - 3y = 11$
Now we will open the bracket by multiplying it by -5:
$ \Rightarrow \dfrac{{ - 75 - 20y}}{2} - 3y = 11$
Multiplying both the sides by 2 we will get:
$ \Rightarrow - 75 - 20y - 6y = 22$
Adding $ - 20y$ & $- 6y$ we will get:
$ \Rightarrow - 75 - 26y = 22$
On rearrangement and addition of like terms together we will get:
$ \Rightarrow - 26y = - 97$
Dividing both sides by $ - 26$:
$ \Rightarrow \dfrac{{ - 26y}}{{ - 26}} = \dfrac{{97}}{{ - 26}}$
$ \Rightarrow y = \dfrac{{ - 97}}{{26}}$
Step3: Substitute $\dfrac{{ - 97}}{{26}}$ for y in the solution to the second equation at the end of step1 and calculate $x$:
$ \Rightarrow x = \dfrac{{15 + 4\left( {\dfrac{{ - 97}}{{26}}} \right)}}{2}$
On solving the bracket we will get:
$ \Rightarrow x = \left( {15 + \dfrac{{ - 97 \times 2}}{{13}}} \right) \div 2$
Taking 13 as L.C.M we will get:
$ \Rightarrow x = \left( {\dfrac{{195 - 194}}{{13}}} \right) \div 2$
On solving the bracket we will get:
$ \Rightarrow x = \dfrac{1}{{13}} \div 2$
On doing the reciprocal of 2 we will get:
$ \Rightarrow x = \dfrac{1}{{13}} \times \dfrac{1}{2}$
On doing the multiplication of fraction we will get:
$ \Rightarrow x = \dfrac{1}{{26}}$

Hence the solution is $x = \dfrac{1}{{26}}:y = \dfrac{{ - 97}}{{26}}$

Note:
This type of question we can solve by two methods: first is substitution and the second one is elimination. In this method the main thing is to find the value of one variable in terms of other students mainly doing the mistakes here.
Alternate method:
We are given two equations i.e.
$ - 5x - 3y = 11$…(1)
$2x - 4y = 15$….(2)
Multiply (2) equation by $5$ and (1) by $2$
$ - 10x - 6y = 22$….(3)
$10x - 20y = 75$….(4)
adding equation (3) and (4) we will get:
$
 {{ - 10x}} - 6y = 22 \\
  \underline {{{10x}} - 20y = 75} \\
   - 26y = 97 \\
 $
Now dividing the both sides by $ - 26$
$y = \dfrac{{ - 97}}{{26}}$
Substitute $y = \dfrac{{ - 97}}{{26}}$ in equation (2) we get the value of x
$ \Rightarrow 2x - 4\left( {\dfrac{{ - 97}}{{26}}} \right) = 15$
On solving the bracket we will get:
$ \Rightarrow 2x + \dfrac{{194}}{{13}} = 15$
Subtracting $\dfrac{{194}}{{13}}$ both sides:
$ \Rightarrow 2x = 15 - \dfrac{{194}}{{13}}$
On subtracting we get:
$ \Rightarrow 2x = \dfrac{{195 - 194}}{{13}}$
 $ \Rightarrow 2x = \dfrac{1}{{13}}$
Dividing both sides by 2 we get:
$ \Rightarrow x = \dfrac{1}{{26}}$
Here also we will get the same solution i.e. $x = \dfrac{1}{{26}};y = \dfrac{{ - 97}}{{26}}$