# 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.

Example

Solve the two equations 2

*x*âˆ’*y*= 1 and 2*x*+ 3*y*= 5.Solution

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

**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 1:***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 2:***Substitute this value in one of the equations to obtain the value of the other variable.**

*Step 3:*Example-1

Solve

3

3

*x*âˆ’ 5*y*= 1*x*+ 4*y*= 6Solution

Given equations are

3

3
3

(3
â‡’ 17

3

3

*x*âˆ’ 5*y*= 1 ....(1)*x*+ 4*y*= 6 ....(2)**Step 1**â€”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.*x*âˆ’ 5*y*= 1 ....(3)*x*+ 12*y*= 18 ....(4)**Step 2**â€”Subtracting (3) from (4), we get*x*+ 12*y*) âˆ’ (3*x*âˆ’ 5*y*) = 18 âˆ’ 1*y*= 17 â‡’*y*= 1**Step 3**â€”Substitute value of*y*in equation 1.*x*âˆ’ 5 Ã— 1 = 1 â‡’*x*= 2**Alternative Method:**

Suppose we have to solve the system of equations

*a*

_{1}

*x*+

*b*

_{1}

*y*=

*c*

_{1}

*a*

_{2}

*x*+

*b*

_{2}

*y*=

*c*

_{2}

_{}

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 | |

a_{1} |
b_{1} |
c_{1} |

a_{2} |
b_{2} |
c_{2} |

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

I | II |

**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

.

**Let us see an example:**

Given equations are

2

3

can be obtained by putting this value of

2

*x*âˆ’ 7*y*= 43

*x*+ 4*y*= 3I | II |

8 âˆ’ (âˆ’21) = 29 | 16 âˆ’ (âˆ’21) = 37 |

can be obtained by putting this value of

*x*in any of the equations [y = (2

*x*âˆ’ 4)/7]

# Substitution Method

Assume that equations are in terms of*x*and

*y*.

**Step 1**â€”Express

*x*in terms of

*y*in one of the equations.

**Step 2**â€”Put the value of

*x*obtained from step 1 in the 2nd equation. Calculate the value of â€˜

*y*â€™ now.

**Step 3**â€”Substitute the value of â€˜

*y*â€™ in any of the equations to find the value of

*x*.

Example

Solve

*x*+ 7*y*= 157

*x*âˆ’ 3*y*= 1Solution

We have

*x*+ 7*y*= 15 ....(1)7

*x*âˆ’ 3*y*= 7 ....(2)From equation (1),

*x*= (15 âˆ’ 7*y*)Substituting this value of

*x*in equation (2),7 Ã— (15 âˆ’ 7

*y*) âˆ’ 3*y*= 1â‡’

*y*= 2 and corresponding value of*x*= 1.