Question

How do you solve the system of equations $2x+2y=4$ and $12-3x=3y$ ?

Answer 1

Hint: In this problem, we rearrange one equation as a function of one variable and put this rearranged equation in the second equation to solve for the other variable. One variable having found out, we can now solve for the other variable. We express the first equation in the form of $y=f\left( x \right)$ and then put this value of $y$ in the second equation to solve for $x$ . then, we solve for $y$ .

Complete step-by-step solution:
The two given equations are
$2x+2y=4....\text{equation }1$
$12-3x=3y....\text{equation }2$
We have two equations and two unknowns, so we can find the solutions of the two equations. We find the solution of the system of linear equations by expressing the first equation in terms of one variable implicitly and then insert this very equation in the next equation. We get an equation containing a single variable, so we can find the value of that variable from here. Putting this value of the variable in another equation, we find the value of the other variable.
It may seem very confusing in words, so we now show it in mathematical form. Rearranging the first equation, we get,
$\Rightarrow 2y=4-2x$
Dividing $2$ on both sides of the equation, we get
$\Rightarrow y=2-x....\text{equation }3$
If we put the value of $y$ from $\text{equation }3$ in $\text{equation }2$ , we get,
\[12-3x=3\left( 2-x \right)....\text{equation }4\]
Applying distributive property to the right hand side of the $\text{equation }4$ , we get
$12-3x=6-3x$
If we observe closely, we can see that the $-3x$ term gets cancelled from both sides of the above equation. This means that $x$ cannot be solved here. We need to analyse why this is so. $\text{equation } 1$ gives $x+y=2$ whereas $\text{equation }2$ gives $x+y=4$ upon simplification. Both the equations can’t be true at the same time as $x+y$ cannot be two different values at the same time.
Therefore, we can conclude that the two given systems of linear equations have no solution.

Note: In these types of problems, students must solve for one variable correctly and must cross-check before putting its value in the other equation. This requires attentiveness. The problem can also be analysed graphically by drawing the respective lines on a graph paper. We will see that the two lines are parallel to each other. This means that there is no point of intersection of the two lines and thus, no solution.