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Graphical Method

We plot the straight lines correspondence to each of the given linear equations in two variables. The point of intersection is the solution for x and y (i.e., two variables) in the system of linear equation.
Solve the two equations 2x  y = 1 and 2x + 3y = 5.
We will draw the graph of both the equations. We can see that the straight lines intersect at the point (1, 1). Hence, the set of solutions of the given equations is (1, 1).

Elimination Method

Step 1: Multiply the co-efficients of the equations by suitable numbers so that the co-efficients of one of the variables becomes same in both the equations.

Step 2: Add or subtract to get one of the variables (which has become same in step 1) cancelled. This will lead to obtaining the value of other variable.

Step 3: Substitute this value in one of the equations to obtain the value of the other variable.
3x − 5y = 1
x + 4y = 6
Given equations are
3x − 5y = 1 ....(1)
x + 4y = 6 ....(2)

Step 1—If we multiply equation (2) by 3, then co-efficient of x will become same in both the equations. Alternatively, if we multiply equation (1) by 4 and equation (2) by 5, co-efficient of y will become same in both the equations. We are making co-efficient of x same in both the equations.
3x − 5y = 1 ....(3)
3x + 12y = 18 ....(4)

Step 2—Subtracting (3) from (4), we get
(3x + 12y) − (3x − 5y) = 18 − 1
⇒ 17 y = 17 ⇒ y = 1

Step 3—Substitute value of y in equation 1.
3x − 5 × 1 = 1 ⇒ x = 2

Alternative Method:
Suppose we have to solve the system of equations
a1x b1y c1
a2x + b2y = c2

Write the co-efficients of x, y and the constants in the following manner (taking notice of the signs, i.e., positive and negative values, of the co-efficients).
I   II
a1 b1 c1
a2 b2 c2

Now, do the cross-multiplication in the part I and part II as given below:

Description: 1259.png
Note that there are two types of cross-multiplication in each of the Part I and Part II - expressed by bold lines and the expressed by dotted lines.

Now, subtract the value obtained by the cross-multiplication of dotted lines from that by the cross-multiplication of bold lines in each of the parts. Suppose the values obtained from part II is m and the value obtained from part I is n; then the value of the variables corresponding to co-efficient Description: 1266.png
Description: 1273.png.
Let us see an example:
Given equations are
2x − 7y = 4
3x + 4y = 3
8 − (−21) = 29 16 − (−21) = 37

can be obtained by putting this value of
 x in any of the equations [y = (2x − 4)/7]

Substitution Method

Assume that equations are in terms of x and y.
Step 1—Express x in terms of y in one of the equations.

Step 2—Put the value of x obtained from step 1 in the 2nd equation. Calculate the value of ‘y’ now.

Step 3—Substitute the value of ‘y’ in any of the equations to find the value of x.
x + 7y = 15
7x − 3y = 1
We have
x + 7y = 15 ....(1)
7x − 3y = 7 ....(2)
From equation (1), x = (15 − 7y)
Substituting this value of x in equation (2),
7 × (15 − 7y) − 3 y = 1
 y = 2 and corresponding value of x = 1.

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