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Question: How do you solve \(3x+y=0\) and \(5x+y=4?\)...

How do you solve 3x+y=03x+y=0 and 5x+y=4?5x+y=4?

Explanation

Solution

As in the given equation, the coefficient of ′x′'x' is 33 in the first equation while in the second equation the coefficient of ′x′'x' is 55.
So, in order to determine the value of the variables ′x′'x' and ′y′'y' multiply equation (i)(i) with 55 and multiply equation (ii)(ii) with 33 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)3x+y=0...(i)
And 5x+y=4...(ii)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′'x' or ′y′'y' in both the equations.
So,
Here let us equalise the coefficient of ′x′'x' in both the equations.
As, coefficient of ′x′'x' in first equation is 33 and coefficient of ′x′'x' in second equation is 5,5,
So,
Multiplying first equation by coefficient of ′x′'x' of second equation i.e. by 5,5, and multiplying second equation by 3,3,
We will get,
5(3x+y=0)5\left( 3x+y=0 \right)
⇒15x+5y=0...(iii)\Rightarrow 15x+5y=0...(iii)
And,
3(5x+y)=4×33\left( 5x+y \right)=4\times 3
⇒15x+3y=12...(iv)\Rightarrow 15x+3y=12...(iv)
Now, subtracting the forth equation from equation (iii)(iii) we will get,
(15x+5y=0)−(15x+3y=12)\left( 15x+5y=0 \right)-\left( 15x+3y=12 \right)
=2y=−12=2y=-12
⇒y=−122=−6\Rightarrow y=\dfrac{-12}{2}=-6
Now, in order to determine the value of ′x′'x' putting the final obtained value of yy in equation (i)(i)
Hence, equation (i)(i) becomes,
3x+y=03x+y=0
⇒3x(−6)=0\Rightarrow 3x\left( -6 \right)=0
⇒3x−6=0\Rightarrow 3x-6=0
⇒3x=6\Rightarrow 3x=6
As, value of 3x3x is 6,6,
So,
Value of xx will be 63=2\dfrac{6}{3}=2

Hence, value of xx and yy will be, (2,−6)\left( 2,-6 \right)

Additional Information:
We can also solve the above given equations by using substitution method,
As, 4x−3y=144x-3y=14
⇒4x=14+3y\Rightarrow 4x=14+3y
⇒4x=14+3y4...(i)\Rightarrow 4x=\dfrac{14+3y}{4}...(i)
Now, putting the value of ′x′'x' in equation (ii)(ii)
y=−3x+4y=-3x+4
y=−3[14+3y4]+4y=-3\left[ \dfrac{14+3y}{4} \right]+4
⇒y=−42−9y4+4\Rightarrow y=\dfrac{-42-9y}{4}+4
⇒y=−42−9y+164\Rightarrow y=\dfrac{-42-9y+16}{4}
⇒y=−26−9y4\Rightarrow y=\dfrac{-26-9y}{4}
⇒4y=−26−9y\Rightarrow 4y=-26-9y
⇒13y=−26\Rightarrow 13y=-26
⇒y=−2613=−2\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,
(2x+3y=1)−(x+2y=2)\left( 2x+3y=1 \right)-\left( x+2y=2 \right)
Here, when we subtract x+2y=2x+2y=2 from 2x+3y=1,2x+3y=1,then coefficient of second equation changed or reversed means is changed to −- and (−)\left( - \right) is changed to (+)\left( + \right)