Question

Do as many as you want
whoever does this will receive brainliest​

Answer 1

Answer:

$\sf\:i\:)\:\boxed{\red{\sf\:k\:=\:2}}\\\\\sf\:ii\:)\:\boxed{\pink{\sf\:k\:=\:4}}\\\\\sf\:iii\:)\:\boxed{\blue{\sf\:k\:=\:4}}\:\sf\:or\:\:\boxed{\red{\sf\:k\:=\:-\:1}}\\\\\sf\:iv\:)\:\boxed{\green{\sf\:k\:=\:7}}$

Step-by-step-explanation:

NOTE : Kindly refer to the attachment first.

We have given that the given equations have infinite solutions.

$\therefore$ We know that, general form of linear equations in two variables is $\sf\:ax\:+\:by\:c\:=\:0$

If equations have infinite solutions, then

$\sf\:\dfrac{a_{1}}{a_{2}}\:=\:\dfrac{b_{1}}{b_{2}}\:=\:\dfrac{c_{1}}{c_{2}}$

By using this property, we can find the value of k.

$\sf\:i\:)\:x\:+\:(\:k\:+\:1\:)y\:=\:5,\:(\:k\:+\:1\:)x\:+\:9y\:=\:8k\:-\:1\\\\\sf\:Comparing\:with\:ax\:+\:by\:+\:c\:=\:0\:,we\:get,\\\\\sf\:a_{1}\:=\:x\:,\:a_{2}\:=\:(\:k\:+\:1\:)\:x\\\\\sf\:b_{1}\:=\:(\:k\:+\:1\:)\:y\:,\:b_{2}\:=\:9y\\\\\sf\:c_{1}\:=\:5\:,\:c_{2}\:=\:8k\:-\:1\\\\\therefore\red{\sf\:\frac{a_{1}}{a_{2}}\:=\:\frac{b_{1}}{b_{2}}\:=\:\frac{c_{1}}{c_{2}}}\\\\\implies\sf\:\frac{a_{1}}{a_{2}}\:=\:\frac{c_{1}}{c_{2}}\\\\\implies\sf\:\frac{x}{(\:k\:+\:1\:)\:x}\:=\:\frac{5}{8k\:-\:1}\\\\\implies\sf\:x\:(\:8k\:-\:1\:)\:=\:5\:(\:k\:+\:1\:)\:x\\\\\implies\sf\:8kx\:-\:x\:=\:5x\:(\:k\:+\:1\:)\\\\\implies\sf\:8kx\:-\:x\:=\:5kx\:+\:5x\\\\\implies\sf\:8kx\:-\:5kx\:=\:5x\:+\:x\\\\\implies\sf\:3k\:\cancel{x}\:=\:6\:\cancel{x}\\\\\implies\sf\:k\:=\:\frac{\cancel6}{\cancel3}\\\\\implies\boxed{\red{\sf\:k\:=\:2}}$

$\sf\:ii\:)\:x\:+\:2y\:+\:7\:=\:0\:,\:2x\:+\:ky\:+\:14\:=\:0\\\\\therefore\sf\:x\:+\:2y\:=\:-\:7\:,\:2x\:+\:ky\:=\:-\:14\\\\\therefore\sf\:a_{1}\:=\:x\:,\:a_{2}\:=\:2x\\\\\sf\:b_{1}\:=\:2y\:,\:b_{2}\:=\:ky\\\\\sf\:c_{1}\:=\:-\:7\:,\:c_{2}\:=\:-\:14\\\\\therefore\red{\sf\:\frac{a_{1}}{a_{2}}\:=\:\frac{b_{1}}{b_{2}}\:=\:\frac{c_{1}}{c_{2}}}\\\\\therefore\sf\:\frac{b_{1}}{b_{2}}\:=\:\frac{c_{1}}{c_{2}}\\\\\implies\sf\:\frac{2y}{ky}\:=\:\frac{\:-\:7}{\:-\:14}\\\\\implies\sf\:2y\:\times\:(\:-\:14\:)\:=\:(\:-\:7\:)\:\times\:ky\\\\\implies\sf\:\cancel{-}\:28y\:=\:\cancel{-}\:7ky\\\\\implies\sf\:28y\:=\:7ky\\\\\implies\sf\:\frac{\cancel{28}\:\cancel{y}}{\cancel{7}\:\cancel{y}}\:=\:k\\\\\implies\boxed{\pink{\sf\:k\:=\:4\:}}$

$\sf\:iii\:)\:5x\:+\:2y\:=\:2k\:,\:2\:(\:k\:+\:1\:)\:x\:+\:ky\:=\:3k\:+\:4\\\\\therefore\sf\:a_{1}\:=\:5x\:,\:a_{2}\:=\:2\:(\:k\:+\:1\:)\:x\\\\\sf\:b_{1}\:=\:2y\:,\:b_{2}\:=\:ky\\\\\sf\:c_{1}\:=\:2k\:,\:c_{2}\:=\:3k\:+\:4\\\\\therefore\red{\sf\:\frac{a_{1}}{a_{2}}\:=\:\frac{b_{1}}{b_{2}}\:=\:\frac{c_{1}}{c_{2}}}\\\\\therefore\sf\:\frac{b_{1}}{b_{2}}\:=\:\frac{c_{1}}{c_{2}}\\\\\implies\sf\:\frac{2\:\cancel{y}}{k\:\cancel{y}}\:=\:\frac{2k}{3k\:+\:4}\\\\\implies\sf\:\frac{2}{k}\:=\:\frac{2k}{3k\:+\:4\:}\\\\\implies\sf\:2\:(\:3k\:+\:4\:)\:=\:2k^{2}\\\\\implies\sf\:6k\:+\:8\:=\:2k^{2}\\\\\implies\sf\:2k^{2}\:-\:6k\:-\:8\:=\:0\\\\\implies\sf\:k^{2}\:-\:3k\:-\:4\:=\:0\:\:\:[\:Dividing\:both\:sides\:by\:2\:]\\\\\implies\sf\:k^{2}\:-\:4k\:+\:k\:-\:4\:=\:0\\\\\implies\sf\:k\:(\:k\:-\:4\:)\:+\:1\:(\:k\:-\:4\:)\:=\:0\\\\\implies\sf\:(\:k\:-\:4\:)\:(\:k\:+\:1\:)\:=\:0\\\\\implies\sf\:k\:-\:4\:=\:0\:\:\:or\:\:\:k\:+\:1\:=\:0\\\\\implies\boxed{\blue{\sf\:k\:=\:4\:}}\:\:\sf\:or\:\:\boxed{\red{\sf\:k\:=\:-\:1\:}}$

$\sf\:iv\:)\:2x\:+\:3y\:=\:7\:,\:(\:k\:-\:1\:)\:x\:+\:(\:k\:+\:2\:)\:y\:=\:3k\\\\\therefore\sf\:a_{1}\:=\:2x\:,\:a_{2}\:=\:(\:k\:-\:1\:)\:x\\\\\sf\:b_{1}\:=\:3y\:,\:b_{2}\:=\:(\:k\:+\:2\:)\:y\\\\\sf\:c_{1}\:=\:7\:,\:c_{2}\:=\:3k\\\\\therefore\red{\sf\:\frac{a_{1}}{a_{2}}\:=\:\frac{b_{1}}{b_{2}}\:=\:\frac{c_{1}}{c_{2}}}\\\\\therefore\sf\:\frac{b_{1}}{b_{2}}\:=\:\frac{c_{1}}{c_{2}}\\\\\implies\sf\:\frac{3y}{(\:k\:+\:2\:)\:y\:}\:=\:\frac{7}{3k}\\\\\implies\sf\:3y\:\times\:3k\:=\:7\:(\:k\:+\:2\:)\:y\\\\\implies\sf\:9ky\:=\:7y\:(\:k\:+\:2\:)\\\\\implies\sf\:9ky\:=\:7ky\:+\:14y\\\\\implies\sf\:9ky\:-\:7ky\:=\:14y\\\\\implies\sf\:2k\:\cancel{y}\:=\:14\:\cancel{y}\\\\\implies\sf\:k\:=\:\frac{\cancel{14}}{\cancel2}\\\\\implies\boxed{\green{\sf\:k\:=\:7}}$