Question

How do you solve the system of equations \[5x+3y=-10\] and \[3x+5y=-6\]?

Answer 1

Hint: From the question given we have been asked to solve \[5x+3y=-10\] and\[3x+5y=-6\]. We can solve the above given equations by using the process of elimination. Here elimination method means we will multiply \[5x+3y=-10\] with 3 and \[3x+5y=-6\] with \[-5\] then by adding these two equations the x term will be cancelled we will get the value of y and by that we will get the value of x. simply we can say that, By using the elimination method, first we will get one variable value and by using that we have to find another variable value.

Complete step by step solution:
From the question given, it has been given that \[5x+3y=-10\]
Let us assume this equation be \[\left( 1 \right)\].
\[3x+5y=-6\]
Let us assume this equation be \[\left( 2 \right)\].
First of all, multiply equation \[\left( 1 \right)\]with 3, we get
\[\Rightarrow 15x+9y=-30\]
Let us assume this equation be \[\left( 3 \right)\]
Now, multiply equation \[\left( 2 \right)\] with \[-5\], we get
\[\Rightarrow -15x-25y=30\]
Let it be equation \[\left( 4 \right)\]
Now, add equation \[\left( 3 \right)\] in equation \[\left( 4 \right)\].
By adding, we get
\[\Rightarrow -16y=0\]
\[\Rightarrow y=0\]
Now, substitute \[y=0\] in the equation \[\left( 3 \right)\].
By substituting \[y=0\] in the equation \[\left( 3 \right)\], we get
\[\Rightarrow 15x+9y=-30\]
\[\Rightarrow 15x+9\times 0=-30\]
\[\Rightarrow 15x=-30\]
\[\Rightarrow x=\dfrac{-30}{15}\]
\[\Rightarrow x=-2\]

Therefore, the solution for the given equations is \[x=-2\] and \[y=0\].

Note: We should be very careful while doing the calculation in this problem. Also, we should know all methods to solve the given equations. we can solve this question by substitution method also. For this question we have chosen an elimination method. Like this, we have to choose their suitable method to solve the given equations.