Solution : 3x + 2y = 11 (1)
2x + 3y = 4 (2)
Let us eliminate 'y' from the given equations. The coefficients of 'y' in the given equations are 2and 3. L.C.M.of 2 and 3 is 6. So, multiply equation (1) by 3 and equation (2) by 2.
Equation (1) × 3
9x + 6y = 33
Equation (2) × 2
4x + 6y = 8(-) (-) (-)
5x = 25x = 255 = 5
Substitute x = 5, in equation (1)
3(5) + 2y = 11
2y = 11 - 15 = - 4 42- Þ = =- y2
Therefore, the required solution is x = 5, y = - 2
.
Do this Solve each of the following pairs of equations by the elimination method.1. 8x+ 5y = 9 2. 2x + 3y = 8 3. 3x + 4y = 253x+2y = 4 4x + 6y = 7 5x - 6y = -9
TRY THISSolve the given pair of linear equations(a - b)x+ (a + b)y = a2 - 2ab - b2
(a + b) (x + y) = a2 + b2
Let us see some more examples:
Example-8. Rubina went to a bank to withdraw `2000. She asked the cashier to give the cashin `50 and `100 notes only. She got 25 notes in all. Can you tell how many notes each of `50and `100 she received?
Solution : Let the number of `50 notes be x;
Let the number of `100 notes be y;
then, x + y = 25
(1)
and 50x + 100y = 2000
(2)
Solutions through the substitution method:
pg:no-93


page no:93

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