Question

How do you solve the system $3x - 2y = 8$ and $2x + 5y = - 1$?

Answer 1

Hint:We are given a pair of linear equations in two variables. There are several methods to solve the system of equations having two variables. Here, we will apply the substitution method. In this method, we express one variable in terms of the other variable in one equation and substitute that value in the next equation to get a linear equation in one variable only. We then solve the linear equation and substitute the value in one equation to get the value of the other variable.

Complete step by step solution:
(i)
We are given two equations:
$
3x - 2y = 8 \\
2x + 5y = - 1 \\
$
These are a pair of linear equations in two variables. In order to solve them, we will apply the substitution method. In this method, we calculate the value of one variable in terms of the other variable and then substitute that value in the second equation.
This will give us a linear equation with one variable which can be solved and when we obtain the value of one variable, we can put it in any of the given equations to obtain the value of another variable.
Therefore, we will calculate the value of any of the two variables from one equation. Since, we have first equation as:
$3x - 2y = 8$
In order to calculate the value of $x$ from this, we will shift everything to RHS except the variable $x$.
Therefore, adding $2y$ both the sides, we will get:
$
3x - 2y + 2y = 8 + 2y \\
3x = 8 + 2y \\
$
Now, we will divide both the sides by $3$. So, we will get:
$
\dfrac{{3x}}{3} = \dfrac{{2y + 8}}{3} \\
x = \dfrac{{2y}}{3} + \dfrac{8}{3} \\
$
(ii)
Since, we have written $x$ in terms of $y$, we will put this value of $x$ in the second equation given to us i.e.,
$2x + 5y = - 1$
Therefore, substituting $x$ as $\dfrac{{2y}}{3} + \dfrac{8}{3}$ in the equation, we will get:
\[2\left( {\dfrac{{2y}}{3} + \dfrac{8}{3}} \right) + 5y = - 1\]
Multiplying both the sides by $3$ and opening the parentheses, we will get:
$4y + 16 + 15y = - 3$
Since, we can add the terms which have the common variable i.e., $y$ with each other, it will become:
$19y + 16 = - 3$
(iii)
Now, since we have got a linear equation in only one variable, we can directly solve it to obtain the value of $y$.
Therefore, in order to solve the equation $19y + 16 = - 3$, we will subtract $16$ from both the sides.
$
19y + 16 - 16 = - 3 - 16 \\
19y = - 19 \\
$
Now we will divide both the sides by $19$:
$\dfrac{{19y}}{{19}} = \dfrac{{ - 19}}{{19}}$
Which will become:
$y = - 1$
(iv)
Now, as we have the value of $y$, we will substitute in the expression we got for $x$ in terms of $y$
i.e., $x = \dfrac{{2y}}{3} + \dfrac{8}{3}$. So, we will get:
$
x = \dfrac{{2\left( { - 1} \right)}}{3} + \dfrac{8}{3} \\
x = \dfrac{{ - 2}}{3} + \dfrac{8}{3} \\
$
Since we have the same denominator in RHS, we can directly add the numerators. Therefore, we will get:
$
x = \dfrac{{ - 2 + 8}}{3} \\
x = \dfrac{6}{3} \\
x = 2 \\
$
Hence, solving the system of equations $3x - 2y = 8$ and $2x + 5y = - 1$ will give us $y = - 1$ and $x = 2$.

Note: Alternate method to solve this system of equation is the elimination method. In this method, we balance any one of the variables to eliminate it by multiplying both the equation with a suitable constant and then adding the equation. For an instance, if we wanted to eliminate $y$, we would have multiplied the first equation $3x - 2y = 8$ with $5$ and the second equation $2x + 5y = - 1$ with $2$ which will provide us with the equations $15x - 10y = 40$ and $4x + 10y = - 2$ respectively.
Now adding these two equations will give us $15x - 10y + 4x + 10y = 40 - 2$ which on simplifying will become $19x = 38$ . From this, we will obtain $x = 2$ and putting the value of $x$ in any of the two equations will give us the value of $y$ i.e., $y = - 1$.