A linear system of two equations with two variables is any system that can be written in the form.
ax+by=p
cx+dy=q
where any of the constants can be zero with the exception that each equation must have at least one variable in it.
Also, the system is called linear if the variables are only to the first power, are only in the numerator and there are no products of variables in any of the equations.
Today we will learn How to solve an equation with two variables.
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Here is an example of a system with numbers.
3x−y=7
2x+3y=1
Before we discuss how to solve systems we should first talk about just what a solution to a system of equations is. A solution to a system of equations is a value of x and a value of y that, when substituted into the equations, satisfies both equations at the same time.
For the example above x=2 and y=−1 is a solution to the system. This is easy enough to check.
3(2)−(−1)=7
2(2)+3(−1)=1
So, sure enough, that pair of numbers is a solution to the system. Do not worry about how we got these values. This will be the very first system that we solve when we get into examples.
Note that it is important that the pair of numbers satisfy both equations. For instance, x=1 and y=−4 will satisfy the first equation, but not the second and so isn’t a solution to the system. Likewise, x=−1 and y=1 will satisfy the second equation but not the first and so can’t be a solution to the system.
x+y=17
x−y=2
For solving these types of equations, we can either go with the substitution method or the elimination method.
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Let’s go with the substitution method first
Given equations are
x+y=12…….(i)
x−y=2……..(ii)
From eq(i) we get,
x+y=12
or, x=12−y…..(iii)
Putting the value of x in eq(ii) we get,
x−y=2
or, 12−y−y=2
or, −2y=2−12
or, −2y=−10
or, y=5y=5[dividing both sides by -2]
Now, putting the value of y in eq(iii) we get,
x=12−y
or, x=12−5
so, x=7
hence, the required solutions are (x,y)=(7,5)
Here, from any one of the equations, we express one variable in terms of another variable [say we expressed x in terms of y]. Then we put this value [of x] in the other equation to find out the value of the other variable [y]. Now we put in the value of this variable [y] in any of the equations to find the value of the variable we started with [x].
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The Second One Is Elimination Method
The elimination method is a bit easier in my opinion. I will take another two equations in this case, but after you understand the process, you can find the solution of the above-mentioned equations too.
Given equations are
2x+3y=22…….(i)
5x−4y=9……..(ii)
Now, multiplying eq(i) by 4 and eq(ii) by 3, we get
8x+12y=88
15x−12y=27
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯[by adding]
23x=115
or, x=115/23
so, x=5
Now, putting the value of x in eq(i) we get,
2x+3y=22
or, 2(5)+3y=22
or, 10+3y=22
or, 3y=22−10
or, 3y=12
so, y=4[dividing both sides by 3]
so, the required solutions are (x,y)=(5,4)
In this method, you multiply both the equations by such a number, so that the coefficients of one variable becomes equal. Here, I’ve multiplied eq(i) by 4 and eq(ii) by 3 and hence made the coefficient of y equal (12).
Then, you’ve to add or subtract in order to eliminate the variable whose coefficient you’ve made equal. I’ve added the two equations here. After adding, you will get an equation of one variable. Find the value of that variable. Then, put in that value in either eq(i) or eq(ii) and find out the value of the other variable.
