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Question: How do you solve \(4x + y = 10\) and \(2x - 3y = 12\)?...

How do you solve 4x+y=104x + y = 10 and 2x−3y=122x - 3y = 12?

Explanation

Solution

To solve a pair of linear equation in two variable, we have to make one of the variables in one of the equation the subject of the equation and then substitute the value of this variable in the other equation, which gives a linear equation in one variable, which is very easy to solve.

Complete step-by-step answer:
Here, we need to find a Cartesian pair (x,y)(x,y) that satisfy both the equations.
Hence, we have to find a unique value of the X coordinate and the Y coordinate that satisfies the conditions of both the equations.
Hence, if we plot these lines on a graph, these lines will intersect at a point (x,y)(x,y) and we need to find that unique point that lies on both lines.
Now, let’s consider the equation of the line 4x+y=104x + y = 10 as an Equation (1)(1) .
And the equation of the line 2x−3y=122x - 3y = 12 as an Equation (2)(2) .
From Equation (1)(1) , let’s make the variable yy as the subject of the equation.
For that subtract 4  x4\;x from both sides of the equation
⇒4x+y−4x=10−4x\Rightarrow 4x + y - 4x = 10 - 4x
⇒y=10−4x\Rightarrow y = 10 - 4x
Consider the above equation as an Equation (3)(3) .
Now, let’s substitute the value of the variable yy from the Equation (3)(3) in the Equation (2)(2) as shown.
⇒2x−3(10−4x)=12\Rightarrow 2x - 3(10 - 4x) = 12
Opening the brackets, we get
⇒2x−3×10+3×4x=12\Rightarrow 2x - 3 \times 10 + 3 \times 4x = 12
⇒2x−30+12x=12\Rightarrow 2x - 30 + 12x = 12
Rearranging the terms,
⇒12x+2x−30=12\Rightarrow 12x + 2x - 30 = 12
Adding   30\;30 on both sides of the equation,
⇒12x+2x−30+30=12+30\Rightarrow 12x + 2x - 30 + 30 = 12 + 30
⇒12x+2x=12+30\Rightarrow 12x + 2x = 12 + 30
Adding the co-efficient of both terms of variable xx and adding the numbers,
⇒14x=42\Rightarrow 14x = 42
Dividing by   14\;14 on both sides,
⇒x=4214\Rightarrow x = \dfrac{{42}}{{14}}
⇒x=3\Rightarrow x = 3
We have got the value of one variable. To find the value of another variable, substituting this value in Equation (3)(3) ,
⇒y=10−4(3)\Rightarrow y = 10 - 4(3)
⇒y=10−12\Rightarrow y = 10 - 12
Hence, the final answer is given as
⇒y=−2\Rightarrow y = - 2
From these two values, the Cartesian point (x,y)(x,y) is given as (3,−2)(3, - 2) .

Hence, the solution of the two equations is (3,−2)(3, - 2).

Note:
Here, any variable from any equation can be taken as the subject of the equation, the answer remains unaffected. To verify the solution we can put the Cartesian point (3,−2)(3, - 2) in the two equations given here and check if they satisfy the equation.