Question

solve the linear equation by elimination method 0.8x+0.5y=0.9;0.6x+0.7y=0.3y​

Answer 1

QUESTION

Solve the linear equation by elimination method

0.8 x + 0.5 y = 0.9

0.6 x + 0.7 y = 0.3

SOLUTION

Given equations

$0.8x+0.5y=0.9$

$\implies \dfrac{8}{10}x + \dfrac{5}{10}y=\dfrac{9}{10}$

Multiplying throughout by 10, we get,

$\implies 8x+5y=9--(1)$

$0.6x+0.7y=0.3$

$\implies \dfrac{6}{10}x + \dfrac{7}{10}y=\dfrac{3}{10}$

Multiplying throughout by 10, we get,

$\implies 6x+7y=3--(2)$

(1) * 3

(2) * 4

$24x+15y=27\\\underline{24x+28y=12}\\\underline{\underline{-13y=15}}$

$\implies y = \dfrac{-13}{15}$

Substituting the value of y in the first equation, we get,

$\implies 8x + 5(\dfrac{-13}{15}) = 9$

$\implies 8x + \dfrac{-13}{3} = 9$

$\implies 8x = 9 + \dfrac{-13}{3}$

$\implies 8x = \dfrac{40}{3}$

$\implies x = \dfrac{5}{3}$

$\boxed{\boxed{\bold{Therefore, \ the \ values \ of \ x \ and \ y \ are \ \frac{5}{3} \ and \ \frac{-13}{15} \ respectivley}}}}}}}$

                                                                                         

Answer 2

GIVEN :-

Equation 1 : 0.8x + 0.5y = 0.9

Equation 2 : 0.6x + 0.7y = 0.3

TO DETERMINE :-

The values of x and y by elimination method.

SOLUTION :-

Multiplying 0.6 to eq.(1) :-

0.6(0.8x + 0.5y) = 0.6(0.9)

⇒0.48x + 0.3y = 0.54          ...(3)

Multiplying 0.8 to eq.(2) :-

0.8(0.6x + 0.7y) = 0.8(0.3)

⇒0.48x + 0.56y = 0.24        ...(4)

Subtracting eq.(4) from eq.(3) :-

$\sf{0.48x + 0.3y = 0.54}\\\sf{-}\\\sf{0.48x + 0.56y = 0.24}\\\rule{90}{1}\\\sf{-0.26y=0.3}$

y = 0.3/-0.26

⇒y = -0.3/0.26

⇒y = -3/10 ÷ 26/100

⇒y = -3/10 × 100/26

⇒y = -30/26

⇒y = -15/13

$\rule{190}{2}$

Now,

from eq.(1), we get :-

x = (0.9 - 0.5y)/0.8

Putting the value of 'y' :-

$\sf{\implies 0.8x + 0.5(\dfrac{-13}{15}) = 0.9}\\\\\\\sf{\implies 0.8x + \dfrac{-13}{30} = 0.9}\\\\\\\sf{\implies 0.8x = 0.9 + \dfrac{13}{30}}\\\\\\\sf{\implies 0.8x = \dfrac{40}{30}}\\\\\\\sf{\implies x = \dfrac{40}{30}\times \dfrac{10}{8}}\\\\\\\sf{\implies x=\dfrac{40}{24} }\\\\\\\boxed{\sf{\implies x=\dfrac{5}{3} }}$

⇒x = 5/3