Question 1 of 9beginner🔧 ApplyShort Answer6 marks

Solve the following pair of linear equations by the substitution method.

(i)

x+y=14x + y = 14 x−y=4x - y = 4

Answer

x = 9, y = 5

(ii)

s−t=3s - t = 3 s3+t2=6\frac{s}{3} + \frac{t}{2} = 6

Answer

s = 9, t = 6

(iii)

3x−y=33x - y = 3 9x−3y=99x - 3y = 9

Answer

Infinitely many solutions

(iv)

0.2x+0.3y=1.30.2x + 0.3y = 1.3 0.4x+0.5y=2.30.4x + 0.5y = 2.3

Answer

x = 2, y = 3

(v)

2x+3y=0\sqrt{2}x + \sqrt{3}y = 0 3x−8y=0\sqrt{3}x - \sqrt{8}y = 0

Answer

x = 0, y = 0

(vi)

3x2−5y3=−2\frac{3x}{2} - \frac{5y}{3} = -2 x3+y2=136\frac{x}{3} + \frac{y}{2} = \frac{13}{6}

Answer

x = 2, y = 3

Explanation

This question from Exercise 3.2 tests the substitution method for solving pairs of linear equations. The method involves expressing one variable in terms of another from one equation, then substituting into the second equation. Part (iii) is a special case where both equations are equivalent, giving infinitely many solutions.

Next