Question

Q1.solve the pair of equations 0.2x + 0. 3y = 1.3 and 0.4x + 0.5y = 2.3 by substitution method .

Answer 1

Step-by-step explanation:

Given:

$\begin{gathered}0.2x + 0.3y = 1.3 \: \: \: ...eq1 \\ 0.4x + 0.5y = 2.3 \: \: \: ...eq2 \\ \end{gathered}$

To find: Solve pair of linear equations by substitution method.

Solution:

Step 1: Simplify the equations by multiplying 10 both sides

$\begin{gathered}2x + 3y = 13 \: \: \: ...eq3\\ \\ 4x + 5y = 23 \: \: \: ...eq4 \\ \end{gathered}$

Step 2: Write eq3 in terms of x and put value of x in eq4

$\begin{gathered}2x = 13 - 3y \\ \end{gathered}$

$\begin{gathered}2(2x) + 5y = 23 \\ \end{gathered}$

put value of 2x

$\begin{gathered}2(13 - 3y) + 5y = 23 \\ \\ 26 - 6y + 5y = 23 \\ \\ - y = 23 - 26 \\ \\ - y = - 3 \\ \\ y = 3 \\ \\ \end{gathered}$

Step 3: Put the value of y in eq 3

$\begin{gathered}2x + 3y = 13 \\ \\ 2x + 3(3) = 13 \\ \\ 2x + 9 = 13 \\ \\ 2x = 13 - 9 \\ \\ 2x = 4 \\ \\ x = \frac{4}{2} \\ \\ x = 2 \\ \\ \end{gathered}$

x=2

Final answer:

Value of x and y are 2 and 3 respectively.

Hope it helps you.

Answer 2

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