Question

solve the equation 0.4 X -1.5 y = 6.5 and 0.3 + 0.2 y = 0.9 by substitution method ​

Answer 1

Given

$0.4x-1.5y=6.5$

$\implies \dfrac{4}{10}x - \dfrac{15}{10}y = \dfrac{65}{10}$

Multiplying by 10, we get,

$\implies 4x-15y=65$

$0.3x+0.2y=0.9$

$\implies \dfrac{3}{10}x + \dfrac{2}{10}y = \dfrac{9}{10}$

Multiplying by 10, we get,

$\implies 3x+2y=9$

$\implies 3x=9-2y$

$\implies x = \dfrac{9-2y}{3}$

Substituting in the first equation, we get,

$\implies 4(\dfrac{9-2y}{3})-15y=65$

$\implies 36-8y-45y=195$

$\implies 36-53y=195$

$\implies -53y=195-36$

$\implies -53y = 159$

$\implies y = -3$

Substituting the value of y, we get,

$x = \dfrac{9+6}{3}= \dfrac{15}{3} = 5$

$\boxed{\boxed{\bold{Therefore, \ the \ values \ of \ x \ and \ y \ are \ 5 \ and \ -3 \ respectively}}}}}}}}}}}$

                                                         

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ values \ of \ x \ and \ y \ are}$

$\sf{5 \ and \ -3 \ respectively.}$

$\sf\orange{Given:}$

$\sf{The \ given \ equations \ are:}$

$\sf{\implies{0.4x-1.5y=6.5}}$

$\sf{\implies{0.3x+0.2y=0.9}}$

$\sf\pink{To \ find:}$

$\sf{Values \ of \ x \ and \ y \ by \ substitution \ method.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ equations \ are:}$

$\sf{\implies{0.4x-1.5y=6.5}}$

$\sf{Multiply \ throughout \ by \ 10}$

$\sf{\implies{4x-15y=65}}$

$\sf{\implies{x=\frac{65+15y}{4}...(1)}}$

$\sf{\implies{0.3x+0.2y=0.9}}$

$\sf{Multiply \ throughout \ by \ 10}$

$\sf{\implies{3x+2y=9...(2)}}$

$\sf{Substitute \ equation (1) \ in \ equation (2)}$

$\sf{3\times\frac{65+15y}{4}+2y=9}$

$\sf{\frac{195}{4}+\frac{45y}{4}+2y=9}$

$\sf{\frac{45y+8y}{4}=9-\frac{195}{4}}$

$\sf{\frac{53y}{4}=\frac{36-195}{4}}$

$\sf{\frac{53y}{4}=\frac{-159}{4}}$

$\sf{Multiply \ both \ sides \ by \ 4}$

$\sf{53y=-159}$

$\sf{\therefore{y=\frac{-159}{53}}}$

$\boxed{\sf{\therefore{y=-3}}}$

$\sf{Substitute \ y=3 \ in \ equation (1), \ we \ get}$

$\sf{x=\frac{65+15(-3)}{4}}$

$\sf{x=\frac{65-45}{4}}$

$\sf{x=\frac{20}{4}}$

$\boxed{\sf{\therefore{x=5}}}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ and \ y \ are}}}$

$\sf\purple{\tt{5 \ and \ -3 \ respectively.}}$