Substitution Method: Algebraic and Graphical Methods

Substitution Method: Algebraic and Graphical Methods | The Complete SAT Course - Class 10 PDF Download

Methods for Solving Linear Equations

Linear equations are a fundamental aspect of algebra and mathematics in general. They are equations where the variables have a degree of 1, and the terms are all constant or linear. When given a system of two linear equations, there are three main methods for solving them algebraically: the substitution method, the elimination method, and the cross-multiplication method.

1. Substitution Method: The substitution method involves solving one equation for one variable and substituting that expression into the other equation. This allows us to find the value of the second variable. The final step is to substitute the value of that variable back into one of the original equations to find the value of the first variable. This method is useful when equations are given in the form of x = ay + b and y = MX + n.
2. Elimination Method: The elimination method involves multiplying or dividing either one or both equations by a number to make the coefficient of one of the variables the same in both equations. Then, we add or subtract the equations to eliminate the variable whose coefficient is the same. Our aim is to find the value of one variable, which can be substituted in either equation to find the other variable. This method is useful when the coefficient of one of the terms is the same in both equations. For example, we can use the elimination method when Ax + By + C = 0 and Px + By + R = 0.
3. Cross-Multiplication Method: The cross-multiplication method involves setting up matching terms in equations before they are combined. This is done through multiplication. Once matching terms are obtained, the equations can be combined, and a solution to the system of linear equations can be found. When using this method, we must multiply all the terms on both sides of the equation, not just the term we are trying to eliminate.

Differences between Substitution and Elimination Methods

Both the substitution and elimination methods are used to solve linear equations algebraically, but there are some differences between them. The elimination method is generally easier to apply when dealing with larger equations or fractions. It involves multiplying or dividing equations by a number to make the coefficient of one variable the same in both equations. In contrast, the substitution method involves finding the value of one variable and substituting it into the other equation to solve for the second variable. This method is often used when equations are given in a specific form, such as x = ay + b and y = MX + n.

What is the Graphical Method?

The graphical method is a technique also referred to as the geometric method used to solve linear equations systems. This method involves developing equations based on constraints and an objective function. To obtain solutions, the graphical method requires several steps. However, in this article, we will focus on explaining in detail how to solve linear equations systems using the first algebraic method, known as the Substitution Method.
Let’s go through the example,
Y = 2x + 4,  3x + y = 9

Now we can substitute y in the second equation with the first equation since we can write the second equation in terms of y(y=y) as
3x + y = 9
y=2x + 4
3x + (2x + 4) =9
5x + 4 =9
5x = 9 - 4
5x = 5
x = 1
This value of x that is equal to 1 can then be used to find the value of y by substituting 1 with x example in the first equation,
Y = 2x + 4y = 2x + 4 = 2 x 1 + 4y
= 2 x 1 + 4 = 6
Y = 6
Therefore, the value of y is equal to 6.
The solution of the linear system is equal to (1, 6).
We can use the substitution method even if both equations of the linear system are in standard form. We can just begin by solving one of the equations for one of its variables.

Questions to be Solved :

Question 1: Solve for the values of ‘x’ and ‘y’:    x + y = 5.   3x + y = 11.

To solve for the values of x and y, we can use the substitution method:
From the first equation, we have:
x + y = 5
y = 5 - x (subtracting x from both sides)
We can substitute this value of y into the second equation:
3x + y = 11
3x + (5 - x) = 11 (substitute y = 5 - x)
2x + 5 = 11
2x = 11 - 5
2x = 6
x = 3
Now that we know the value of x, we can substitute it into one of the original equations to solve for y. Let's use the first equation:
x + y = 5
3 + y = 5 (substitute x = 3)
y = 2
Therefore, the solution for the values of x and y is:
x = 3, y = 2.

Question 2: Solve for the values of ‘x’ and ‘y’:       2x + 6y = 10,       1x - 2y = 15

To solve for the values of 'x' and 'y' in the equations:
2x + 6y = 10
1x - 2y = 15
We can use the substitution method or the elimination method. Here's how to solve it using the substitution method:
Solve one of the equations for one variable (in terms of the other variable):
Let's solve the second equation for 'x':
1x - 2y = 15
1x = 2y + 15
Substitute the expression found in step 1 into the other equation:
2(2y + 15) + 6y = 10
4y + 30 + 6y = 10
10y = -20
y = -2
Substitute the value of 'y' back into one of the original equations to solve for 'x':
2x + 6y = 10
2x + 6(-2) = 10
2x - 12 = 10
2x = 22
x = 11
Therefore, the solution for the system of equations is x = 11 and y = -2.

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