Question

5x + 4y = 20 and 3x + 2y = 12​

Answer 1

Question:-

To solve the pair of linear equation:-

5x + 4y = 20 and 3x + 2y = 12

Method:-

Substitution Method

Solution:-

$\sf{5x+4y=20 \longrightarrow (i)}$

$\sf{3x+2y=12 \longrightarrow (ii)}$

From eq.(i)

$\sf{5x + 4y = 20}$

$\sf{\implies 5x = 20-4y}$

$\sf{\implies x = \dfrac{20-4y}{5}}$

Substituting the value of x in eq.(ii)

$\sf{3x + 2y = 12}$

= $\sf{3\times\bigg(\dfrac{20-4y}{5}\bigg) + 2y = 12}$

$\sf{\implies \dfrac{60-12y}{5} + 2y = 12}$

$\sf{\implies \dfrac{60-12y+10y}{5} = 12}$

$\sf{\implies 60 - 2y = 12\times 5}$

$\sf{\implies 60 - 2y = 60}$

$\sf{\implies -2y = 60-60}$

$\sf{\implies -2y = 0}$

$\sf{\implies y = \dfrac{0}{-2}}$

$\sf{\implies y = 0}$

Putting the value of y in eq.(i)

$\sf{5x + 4y = 20}$

= $\sf{5x + 4\times 0 = 20}$

$\sf{\implies 5x + 0 = 20}$

$\sf{\implies 5x = 20}$

$\sf{\implies x = \dfrac{20}{5}}$

$\sf{\implies x = 4}$

$\sf{\therefore The \:value \:of \:x = 4 \:and\:the\:value\:of\:y=0}$