What is the Substitution Method in Math?
The method of solving simultaneous equations by putting the value of one variable in terms of another variable from one equation to another equation is called the Substitution Method in Math.
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What are Simultaneous Equations? What is mean by Solving Simultaneous Equations?
Simultaneous Equations are a pair of equations having a unique solution that satisfies both equations. And, Solving Simultaneous Equations means finding that unique solution.
For example: x
+ y = 5 and x – y = 3 are a pair of simultaneous linear equations. And, x = 4,
y = 1 are the unique solutions that satisfy both the equations.
Read Simultaneous Linear Equations for more details.
How to Solve by Substitution Method?
Here are the STEPS for
Substitution METHOD:
Step 1: From anyone equation, find the value of one variable in terms
of another variable.
Step 2: Substitute that value in another equation.
Step 3: Then, simplify that equation, this will give the numerical value
of one variable.
Step 4: Substitute that value in any equation and get the value of
another variable.
Things To Remember:

Solve any one equation
either for x or y.
 Then, substitute the value of x or y in the next equation and solve it.
Here is an example of Substitution
Method:
Example 1: Solve the following
simultaneous equations by substitution method: x + y = 5 and x – y = 3.
Solution:
Here, the given simultaneous equations are:
x + y = 5 …………… (i)
x – y = 3 …………… (ii)
From equation (I),
x + y = 5
or, x = 5 – y ……… (iii)
Now, substituting the value of x = 5 – y from equation (iii) to equation (ii) we get,
5 – y – y = 3
or, 5 – 2y = 3
or, – 2y = 3 – 5
or, – 2y = – 2
or, y = – 2/– 2
or, y = 1
Again, substituting the value of y = 1 in equation (i) we get,
x + 1 = 5
or, x = 5 – 1
or, x = 4
∴ x = 4 and y = 1.
Answer.
In this way, we can solve the system of simultaneous linear equations by substitution method.
Here are some more examples of substitution method.
Worked Out Examples
Example 2: Solve by subtitution method: x = 3y and 4x + y = 26.
Solution: Here, the given simultaneous equations are:
x = 3y ………… (i)
4x + y = 26 ………… (ii)
Substituting the value of x = 3y from equation (i) to equation (ii) we get,
4 × 3y + y = 26
or, 12y + y = 26
or, 13y = 26
or, y = 26/13
or, y = 2
Again, substituting the value of y = 2 in equation (i) we get,
x = 3 × 2
or, x = 6
Hence, x = 6 and y = 2 Answer.
Example 3: Solve by subtitution method: 2x + y = 180 and x + 2y = 240.
Solution: Here, the given simultaneous equations are:
2x + y = 180 ……………… (i)
x + 2y = 240 ……………… (ii)
From equation (i),
y = 180 – 2x ……………… (iii)
Now, substituting the value of y 180 – 2x from equation (iii) to equation (ii), we get,
x + 2 (180 – 2x) = 240
or, x + 360 – 4x = 240
or, – 3x = 240 – 360
or, – 3x = – 120
or, x = – 120 / –3
or, x = 40
Again, substituting the value of x = 40 in equation (iii) we get
y = 180 – 2 × 40
= 180 – 80
= 100
Hence, x = 40 and y = 100 Answer.
Example 4: Solve by subtitution method: 3x – 2y = 5 and 7x + 3y = 27.
Solution: Here, the given simultaneous equations are:
3x – 2y = 5 …………… (i)
7x + 3y = 27 …………… (ii)
From equation (i),
3x = 5 + 2y
or, x = (5 + 2y)/3 …… (iii)
Now, substituting the value of x = (5 + 2y)/3 from equation (iii) to equation (ii) we get,
7×(5 + 2y)/3 + 3y = 27
or, [7(5 + 2y) + 9y]/3=
27
or, 35 +14y + 9y = 27 ×
3
or, 23y = 81 – 35
or, y = 46/23
or, y = 2
Again, substituting the value of y = 2 in equation (iii) we get,
x = (5+2×2)/3
or, x = 9/3
or, x = 3
Hence, x = 3 and y = 2 Answer.
If you have any questions or problems regarding the Substitution Method, you can ask here, in the comment section below.
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