Question

How do you solve $5x - 3y = 16$ and $4x + 5y = - 2$?

Answer 1

Hint: In this question, we are given two equations and we have been asked to solve them. You can use any method out of the three methods – elimination, substitution or cross multiplication. I will solve this question using two methods: elimination and substitution.
Substitution method: The two equations given are in terms of $x$ and $y$. Using one equation, find the value of one variable. Then, put this in the other equation such that the equation is in one variable only. Then, using this equation, find the value of that variable. Put this in any of the original equations and find the value of the other variable.

Complete step-by-step solution:
Let us solve by substitution method first. We have two equations –
$ \Rightarrow 5x - 3y = 16$ ……... (1)
Rearranging the equation in such a way that we get the value of $x$,
$ \Rightarrow x = \dfrac{{16 + 3y}}{5}$ …..…. (2)
$ \Rightarrow 4x + 5y = - 2$ ……... (3)
Put equation (2) in equation (3),
$ \Rightarrow 4\left( {\dfrac{{16 + 3y}}{5}} \right) + 5y = - 2$
Simplifying the equation by taking a common denominator,
$ \Rightarrow 4\left( {16 + 3y} \right) + 25y = - 10$
Opening the brackets,
$ \Rightarrow 64 + 12y + 25y = - 10$
Shifting to find the value,
$ \Rightarrow 37y = - 74$
$ \Rightarrow y = \dfrac{{ - 74}}{{37}} = - 2$
Now, putting this in equation (3) to find the value of $x$,
$ \Rightarrow 4x + 5\left( { - 2} \right) = - 2$
$ \Rightarrow 4x = - 2 + 10$
$ \Rightarrow x = \dfrac{8}{4} = 2$

Hence, $x = 2,y = - 2$.

Note: Alternative method:
Elimination method: In this method, we eliminate one variable and find the value of the other. Choose any one variable. Make their coefficients same by multiplying or dividing a constant in the entire equation. Once it is done, subtract the equations. This will eliminate that variable which you chose to make the same. Find the value of the other variable. Put this in any of the original equations and find the value of the other variable.
Now, we will solve the same question using the elimination method.
$ \Rightarrow 5x - 3y = 16$ …. (1)
$ \Rightarrow 4x + 5y = - 2$ …. (2)
Now, if we look at the equation (1), $y$ has a variable $3$ and in equation (2), $y$ has a variable $5$. So, we will multiply equation (2) by 3 and equation (1) by $5$ so as to make the coefficients the same.
$ \Rightarrow 5\left( {5x - 3y = 16} \right) \Rightarrow 25x - 15y = 80$
$ \Rightarrow 3\left( {4x + 5y = - 2} \right) \Rightarrow 12x + 15y = - 6$
Now, we will add both the final equations,
$ \Rightarrow 25x - 15y + 12x + 15y = 80 - 6$
Now, our variable $y$ will be eliminated.
$ \Rightarrow 25x + 12x = 80 - 6$
$ \Rightarrow 37x = 74$
$ \Rightarrow x = 2$
Putting in equation (2),
$ \Rightarrow 4\left( 2 \right) + 5y = - 2$
Simplifying,
$ \Rightarrow 5y = - 2 - 8 = - 10$
$ \Rightarrow y = - 2$
Hence, $x = 2,y = - 2$.