Question

Solve the following pair of linear equations by the elimination method :
0.5x+2y/3=1 and X-y/3=0​

Answer 1

Given:

• 0.5x + 2y/3 = 1
• x - y/3 = 0

Find:

• Solve the given pair of Linear Equation

Solution:

we, have

$\to \sf 0.5x + \dfrac{2y}{3} = (1).......1$

$\to \sf x - \dfrac{y}{3} = 0........(2)$

Multiplyimg Eq 2 by 0.5 and Eq 1 by 1

$\to \sf 0.5x + \dfrac{2y}{3} = 1... \times 1$

$\to \sf x - \dfrac{y}{3} = 0..... \times 0.5$

we, got

$\to \sf 0.5x + \dfrac{2y}{3} = 1.........(3)$

$\to \sf 0.5x - \dfrac{0.5y}{3} = 0..........(4)$

Substract eq 4 from 3

$\sf 0.5x + \dfrac{2y}{3} = 1 \\ \underline{\sf - \boxed{ + } 0.5x + \boxed{-} \dfrac{0.5y}{3} = 0} \\ \sf \dfrac{2y}{3} + \dfrac{0.5y}{3} = 1 \\ \sf \dfrac{2y + 0.5y}{3} = 1 \\ \sf 2.5y = 3 \\ \sf y = \dfrac{3}{2.5} = 1.2$

Substitute value of y in eq(1) we, get

$\to \sf 0.5x + \dfrac{2y}{3} = 1$

$\to \sf 0.5x + \dfrac{2(1.2)}{3} = 1$

$\to \sf 0.5x + \dfrac{2.4}{3} = 1$

$\to \sf 0.5x = 1 - \dfrac{2.4}{3}$

$\to \sf 0.5x = \dfrac{3 - 2.4}{3}$

$\to \sf 0.5x = \dfrac{0.6}{3}$

$\to \sf 0.5x = 0.2$

$\to \sf x = \dfrac{0.2}{0.5}$

$\to \sf x = 0.4$

Hence, x = 0.4 and y = 1.2