Find out which of the following equations have x=1, y=2 as solution.
x + 2y = 5
2y - 3x + 3 = 0
3x + 4y - 8 = 0 [4 MARKS]

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Solution

Concept and result : 1 Mark
Equation calculation : 1 Mark each

Putting x = 1, y = 2 in the equation x + 2y = 5,

1 + (2x2) = 5

The equation holds. Hence x = 1, y = 2 is a solution of the equation x + 2y = 5

Putting x = 1, y = 2 in the equation 2y – 3x + 3 =0,

$(2 \times 2) - (3 \times 1) + 3 = 4 \neq 0$

The equation does not hold. Hence x = 1, y = 2 is not a solution of the equation 2y – 3x + 3 =0

Putting x = 1, y = 2 in the equation 3x + 4y – 8 =0,

$(3 \times 1) + (4 \times 2) - 8 = 3 \neq 0$

The equation does not hold. Hence x = 1, y = 2 is not a solution of the equation 3x + 4y – 8 =0

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Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x+y=5 and 2x−3y=4
(ii) 3x+4y=10 and 2x−2y=2
(iii) 3x−5y−4=0 and 9x=2y+7
(iv) x2+2y3=−1 and x−y3=3

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