Question

How do you solve the system \[2x+y=9\] and \[8x-2y=6\] ?

Answer 1

Hint: In this question, we have to solve two linear equations, to get the value of x and y. So, we will use the Substitution method to solve the problem. We first rewrite the first equation in terms of x and then put it into the second equation, and solve the variable x. therefore, on further calculations, we see that both the variables cancel out and we are left with the constants only, which implies that the system of equations is inconsistent, which is our required answer to the question.

Complete step-by-step solution:
According to the question, we have to find the values of x and y.
Therefore, we will use the substitution method to do the same.
The equations given in the problem are: \[2x+y=9\] ---------- (1) and \[8x-2y=6\] ------- (2)
So, we will first rewrite the equation (1) in terms of x, which is
\[2x+y=9\]
Now, subtract 2x on both sides of the above equation, we get
\[\Rightarrow 2x+y-2x=9-2x\]
As we know, the same terms with opposite signs cancel out each other, therefore we get
\[\Rightarrow y=9-2x\] -------- (3)
Now, we will put the value of equation (3) into equation (2), we get
\[\Rightarrow 8x-2(9-2x)=6\]
Now, we will open the brackets of the above equation, we get
\[\Rightarrow 8x-18+4x=6\]
As we know, the same terms with opposite signs cancel out each other, therefore we get
\[\begin{align}
  & \Rightarrow 12x=24 \\
 & \Rightarrow x=2 \\
\end{align}\]
Thus, we see that in the above equation, the variable x has value 2. Therefore, we will find the value of y. From equation (3), we get
\[\begin{align}
  & y=9-2x \\
 & \Rightarrow y=9-2(2) \\
 & \Rightarrow y=9-4 \\
 & \Rightarrow y=5 \\
\end{align}\]
Therefore, the solution of the above equation is x=2, y=5.

Note: While solving this problem, do mention all the methods you are using to avoid confusion and mathematical error. One of the alternative methods to solve this problem is using the cross-multiplication method; we will multiply the variable of the numerator of each side by the denominator of the other side, to get the required result of the problem, which is no solution for the problem.