Math, asked by Tarunvirat, 11 months ago

Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:

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Answers

Answered by kuldeepmaurya70
1

Step-by-step explanation:

y \div 15 + y \div 12 = 4 - 19 \div 8 \\ 12y + 15y = 13  \times 15 \times 12 \div 8 \\ 27y = 13 \times 45 \ \div 2 \\ 54y = 585 \\ y = 585 \div 54 \\ y = 585 \div 54 \\

Answered by Anonymous
3

 \frac{x}{6}  +  \frac{y}{15}  = 4 \\  =  > 5x + 2y = 120 \\  =  > 5x = 120 - 2y \\  =  > 20x = 480 - 8y \:  \:  \\ and \\  \frac{x}{3}  -  \frac{y}{12}  = 4 \frac{3}{4}  \\  =  >  \frac{4x - y}{12}  =  \frac{19}{4}  \\  =  > 4x - y = 57 \\  =  > 20x - 5y = 285 \\ puttin g \: the \: value \: of \: 20x \:  \: we \: get \:  \:  \\  =  > 480 - 8y - 5y = 285 \\  =  >  13y =  480  - 285\\  =  > y =  \frac{  195 }{ 13}  \\  =  > y = 15 \\  \\ therefore \\ 5x = 120 - 2(15) \\  =  > x =  \frac{90}{5}  \\  =  > x = 16

therefore x=16 and y=15

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