Question

How do you solve $3x+y=0$ and $5x+y=4?$

Answer 1

Hint: As in the given equation, the coefficient of $'x'$ is $3$ in the first equation while in the second equation the coefficient of $'x'$ is $5$.
So, in order to determine the value of the variables $'x'$ and $'y'$ multiply equation $(i)$ with $5$ and multiply equation $(ii)$ with $3$ and then by subtracting the equations determine the values of variables.

Complete step-by-step answer:
As per data given in the question,
We have,
$3x+y=0...(i)$
And $5x+y=4...(ii)$
Here, both given equations are a type of linear equation of two variables. So, in order to determine the value of variables, we will equalise the coefficient of either $'x'$ or $'y'$ in both the equations.
So,
Here let us equalise the coefficient of $'x'$ in both the equations.
As, coefficient of $'x'$ in first equation is $3$ and coefficient of $'x'$ in second equation is $5,$
So,
Multiplying first equation by coefficient of $'x'$ of second equation i.e. by $5,$ and multiplying second equation by $3,$
We will get,
$5\left( 3x+y=0 \right)$
$\Rightarrow 15x+5y=0...(iii)$
And,
$3\left( 5x+y \right)=4\times 3$
$\Rightarrow 15x+3y=12...(iv)$
Now, subtracting the forth equation from equation $(iii)$ we will get,
$\left( 15x+5y=0 \right)-\left( 15x+3y=12 \right)$
$=2y=-12$
$\Rightarrow y=\dfrac{-12}{2}=-6$
Now, in order to determine the value of $'x'$ putting the final obtained value of $y$ in equation $(i)$
Hence, equation $(i)$ becomes,
$3x+y=0$
$\Rightarrow 3x\left( -6 \right)=0$
$\Rightarrow 3x-6=0$
$\Rightarrow 3x=6$
As, value of $3x$ is $6,$
So,
Value of $x$ will be $\dfrac{6}{3}=2$

Hence, value of $x$ and $y$ will be, $\left( 2,-6 \right)$

Additional Information:
We can also solve the above given equations by using substitution method,
As, $4x-3y=14$
$\Rightarrow 4x=14+3y$
$\Rightarrow 4x=\dfrac{14+3y}{4}...(i)$
Now, putting the value of $'x'$ in equation $(ii)$
$y=-3x+4$
$y=-3\left[ \dfrac{14+3y}{4} \right]+4$
$\Rightarrow y=\dfrac{-42-9y}{4}+4$
$\Rightarrow y=\dfrac{-42-9y+16}{4}$
$\Rightarrow y=\dfrac{-26-9y}{4}$
$\Rightarrow 4y=-26-9y$
$\Rightarrow 13y=-26$
$\Rightarrow y=\dfrac{-26}{13}=-2$

Note:
When we subtract one equation from other, then the coefficient of equation which are subtracted being reversed like,
$\left( 2x+3y=1 \right)-\left( x+2y=2 \right)$
Here, when we subtract $x+2y=2$ from $2x+3y=1,$then coefficient of second equation changed or reversed means is changed to $-$ and $\left( - \right)$ is changed to $\left( + \right)$