Question

If the system of equations 2x + 3y = 7 and (a+b)x + (2a-b)y = 21 has infinitely many solutions, then a) a = 1, b = 5 b) a = 5, b = 1 c) a = -1, b = 5 d) a = 5, b = -1 pls ans fast urgent problem ....................

Answer 1

Answer:

Therefore a = 5 , b= 1

Therefore option B

Step-by-step explanation:

$Equation \ 1 =2x+3y-7\\Equation \ 2 =(a+b)x + (2a-b)y-21\\For \ infinite \ solutions\\\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}\\\frac{2}{a+b} = \frac{3}{2a-b} = \frac{-7}{-21}$

$\frac{2}{a+b} = \frac{-7}{-21}\\\frac{2}{a+b} = \frac{1}{3}\\6=a+b\\a+b=6...(1)$

$\frac{3}{2a-b} = \frac{-7}{-21}\\\frac{3}{2a-b} = \frac{1}{3}\\9=2a-b\\2a-b=9...(2)$

$(1)+(2)\\3a=15\\a=\frac{15}{3} = 5\\in \ (1)\\a+b=6\\5+b=6\\b=6-5=1$

Therefore a = 5 , b= 1

Therefore option B

Answer 2

Answer:

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Step-by-step explanation: