Question

solve the following pair of linear equation by the elimination method and substitute method x/4+y/3=1 and 5x+3y=2​

Answer 1

Answer:

x = -28/11

y = 54/11

Step-by-step explanation:

Using Elimination method,

$\frac{x}{4} +\frac{y}{3} =1\\\\\frac{3x+4y}{12} =1\\\\(3x+4y=12)*3----(i)\\\\9x+12y=36------(ii)$

$(5x + 3y = 2)*4-----(iii)\\\\20x+12y=8----(iv)$

Subtract (ii) - (iv)

9x + 12y - (20x + 12y) = 36 - 8

9x  + 12y - 20x - 12y = 28

-11x = 28

x = $\frac{-28}{11}$

20x + 12y = 8

$20(\frac{-28}{11} ) + 12y = 8\\\\\frac{-560}{11} +12y=8\\\\12y=8+\frac{560}{11} \\\\12y=\frac{88+560}{11} \\\\y=\frac{648}{132} \\\\y=\frac{54}{11}$

Using Substitution method,

$\frac{x}{4} +\frac{y}{3} =1\\\\\frac{3x+4y}{12} =1\\\\3x+4y=12\\\\3x={12-4y} \\\\x=\frac{12-4y}{3} ----(i)$

$5x+3y=2\\\\5(\frac{12-4y}{3} )+3y=2\\\\\frac{60x-20y}{3} +3y=2\\\\\frac{60-20y+9y}{3} =2\\\\60-20y+9y=6\\\\60-11y=6\\\\11y=60-6\\\\11y=54\\\\y=\frac{54}{11}$

$x=\frac{12-4y}{3} \\\\3x=12-4(\frac{54}{11} )\\\\3x=12-\frac{216}{11} \\\\3x=\frac{132-216}{11} \\\\x=\frac{-84}{33} \\\\x=\frac{-28}{11}$