Question

How do you solve the system \[x+y=4\] and \[2x+3y=0\] ?

Answer 1

Hint: These types of problems are pretty straight forward and are very easy to solve. For such problems, we need to remember about the theory of linear as well as simultaneous equations. The above two mentioned equations are that of a pair of simultaneous equations which are generally used to solve for the unknown parameters \[x\] and \[y\] . These two set of linear equations are independent of each other and once solved gives a real value of \[x\] and \[y\]. For solving we need to multiply the first equation with an integer as well as the second equation with an integer in such a way that on adding or subtracting one of the two unknown parameters get eliminated and we can very easily find the value of the other. Putting this known parameter value in the original equation, we can find the value of the other.

Complete step by step solution:
Now we start off with the solution to the problem by writing the two equations as,
\[x+y=4\] ---- Equation \[1\]
\[2x+3y=0\] ----- Equation \[2\]
We know multiply the first equation with \[3\] on both the sides and it transforms to,
\[3x+3y=12\] ---- Equation \[3\]
We keep the second equation as it is, or we may say that we multiply the second equation by \[1\] on both the sides.
We now subtract the second equation from the third and we get,
\[3x+3y-2x-3y=12\]
On evaluating we get,
\[\begin{align}
  & \Rightarrow 3x-2x=12 \\
 & \Rightarrow x=12 \\
\end{align}\]
Now, putting this value of \[x\] in equation \[1\] we find the value of \[y\] as,
\[\begin{align}
  & 12+y=4 \\
 & \Rightarrow y=-8 \\
\end{align}\]

Thus the solution to our problem is \[x=12\] and \[y=-8\]

Note: For these types of problems we need to remember the theory of equations chapter very well. The given problem can also be solved graphically. We need to plot the given two straight lines and find out the point of intersection of the two straight lines. The intersecting point gives the required value of ‘x’ and ‘y’. We need to be very careful and observant in making a choice of the integers that we have to multiply in both the given equations.