Question

How do you substitute methods to solve $ 3x - 15y = - 12 $ and $ 3x + 24y = - 12 $ ?

Answer 1

Hint: In the given question, we need to solve two simultaneous equations in two variables. There are various methods to solve two given equations in two variables like substitution method, cross multiplication method, elimination method, matrix method and many more. The equations given in the question can be solved using any one of the above mentioned methods easily. But we will solve the equations using the substitution method as mentioned in the question.

Complete step-by-step answer:
In the question, we are given a couple of simultaneous linear equations in two variables.
 $ 3x - 15y = - 12 - - - - - - - \left( 1 \right) $
 $ 3x + 24y = - 12 - - - - - - - \left( 2 \right) $
In the substitution method, we substitute the value of one variable from an equation into another equation so as to get an equation in only one variable.
So, from first equation, we have,
 $ \Rightarrow 3x = 15y - 12 $
 $ \Rightarrow x = \left( {\dfrac{{15y - 12}}{3}} \right) $
 $ \Rightarrow x = 5y - 4 $
Now putting the value of y obtained from one equation into another, we get,
 $ 3\left( {5y - 4} \right) + 24y = - 12 $
 $ 15y - 12 + 24y = - 12 $
 $ 39y = - 12 + 12 $
 $ 39y = 0 $
 $ y = 0 $
So, we get, $ y = 0 $
Putting the value of y in any of the two equations to find x, we get,
 $ 3x - 15\left( 0 \right) = - 12 $
 $ \Rightarrow x = \dfrac{{ - 12}}{3} $
 $ \Rightarrow x = - 4 $
Therefore ,the solution of the simultaneous linear equations $ 3x - 15y = - 12 $ and $ 3x + 24y = - 12 $ is $ x = - 4 $ and $ y = 0 $ .
So, the correct answer is “ $ x = - 4 $ and $ y = 0 $ ”.

Note: Linear Equation in two variables: A equation consisting of 2 variables having degree one is known as Linear Equation in two variables. Standard form of Linear Equation in two variables is $ ax + by + c = 0 $ where a, b and c are the real numbers and a, b which are coefficients of x and y respectively are not equal to 0.