Question

How do you solve the system \[y = 6x - 11\] and \[ - 2x - 3y = - 7\] by substitution?

Answer 1

Hint: Here, we will first substitute the value of any of the one variable from either of the equations in the other equation. We will then solve that equation having a single variable to find the value of that specific variable. Again, by substituting the value of the obtained variable in any equation, we will be able to find the values of both variables.

Complete step by step solution:
The given linear equations in two variables are:
\[y = 6x - 11\]………………………….\[\left( 1 \right)\]
\[ - 2x - 3y = - 7\]………………………\[\left( 2 \right)\]
Now, substituting the value of \[y\] from equation \[\left( 1 \right)\] in \[\left( 2 \right)\], we get
\[ - 2x - 3\left( {6x - 11} \right) = - 7\]
Using distributive property of multiplication, we get
\[ \Rightarrow - 2x - 18x + 33 = - 7\]
Adding the like terms, we get
\[ \Rightarrow - 20x + 33 = - 7\]
Adding 7 on both sides, we get
\[ \Rightarrow - 20x + 33 + 7 = 0\]
Adding \[20x\] on both the sides, we get
\[ \Rightarrow 40 = 20x\]
Dividing both sides by 20, we get
\[ \Rightarrow x = 2\]
Substituting the value of \[x\] as 2 in equation \[\left( 1 \right)\], we get,
\[y = 6\left( 2 \right) - 11 = 12 - 11 = 1\]
Therefore, we have solved the given equations using the substitution method.

Thus, the solution of the given linear equations in two variables is \[\left( {x,y} \right) = \left( {2,1} \right)\]

Note:
An equation is called linear equation in two variables if it can be written in the form of \[ax + by + c = 0\] where \[a,b,c\] are real numbers and \[a \ne 0\] , \[b \ne 0\] as they are coefficients of \[x\] and \[y\] respectively. Also, the power of linear equations in two variables will be 1 as it is a ‘linear equation’. Also, a linear equation in two variables can sometimes have infinitely many solutions rather than only one in the case of ‘one variable’.
Now, when we open a bracket by multiplying each term by a number that is present outside the bracket, we need to keep in mind that since there is a negative sign outside the bracket. Thus, we will change the minus sign present inside the bracket to its opposite sign, i.e. plus.