Question

How do solve the following linear system: $x+2y=1,x+y=6$ ?

Answer 1

Hint: We will use the Elimination Method to solve the given linear systems $x+2y=1$ and $x+y=6$. Since, in the given linear systems, the coefficient of $x$ is the same for both the equations that is $1$ . According to the elimination method, we don’t need to use the step in that we have to make the coefficient of one variable the same. Since, the sign of the coefficients are the same, is a positive sign, we will subtract one equation from the other equation. From the above step, we will be able to cancel out the $x$ term and will get the value of $y$. Now, we can use this value of $y$ by substituting in any equation to get the value of $x$. After solving the equation with help of the value of $y$, we will find the value of $x$ .

Complete step by step solution:
Here are the linear equations as:
\[\Rightarrow x+2y=1\] … $\left( i \right)$
And
\[\Rightarrow x+y=6\]. … $\left( ii \right)$
Since, the coefficient of $x$ is common in both equations, so there is no need to multiply or divide with any number in the above equations.
Now, we will subtract equation $\left( ii \right)$ from equation $\left( i \right)$ to cancel out the $x$ term as:
\[\Rightarrow \left( x+2y \right)-\left( x+y \right)=1-6\]
After opening the bracket, the signs of second bracketed terms will be changed after multiplying the negative sign that is outside of the bracket. Then, the above equation will be as:
 \[\Rightarrow x+2y-x-y=1-6\]
Here, we will look for equal like terms and will combine them as:
\[\Rightarrow \left( x-x \right)+\left( 2y-y \right)=\left( 1-6 \right)\]
By subtracting $x$ from $x$, the $x$ term will be eliminated. Similarly subtracting $y$ from $2y$, we will get $y$and \[\left( 1-6 \right)\] is equal to $\left( -5 \right)$ from the above step and the next step will be as:
\[\Rightarrow y=-5\]
Here, we had the value of $y$ as:
\[\Rightarrow y=-5\]
Now, for getting the value of $x$, we can substitute the value for $y$ in any above equation. After putting this value in equation $\left( ii \right)$ , the equation will be:
\[\Rightarrow x+y=6\]
\[\Rightarrow x+\left( -5 \right)=6\]
Since, the multiplication of a negative sign and a positive sign is equal to negative sign.
\[\Rightarrow x-5=6\]
Now, we will have to place all the number one side as:
\[\Rightarrow x=6+5\]
 Here, we will complete the calculation in the above equation. After adding, we got the value of $x$ :
\[\Rightarrow x=11\]
Hence, we had a result in which the values are \[x=11\] and \[y=-5\].

Note: Since, we got the values \[11\] and \[-5\] for $x$ and $y$ respectively. We can verify it by putting these values in any given linear equations. Let’s select an equation as:
\[\Rightarrow x+y=6\]
Now, we will put the value of $x$ and $y$ respectively in L.H.S. as:
L.H.S.:
\[\Rightarrow 11+\left( -5 \right)\]
\[\Rightarrow 11-5\]
\[\Rightarrow 6\]
That is R.H.S of the selected equation. Hence, the solution is correct for the given question.