Question

find the value of kfor which the system
2x+KY=1 and 3x-5y=7

Answer 1

$\huge\blue{\mid{\fbox{\tt{Solution}}\mid}}$

$\bold{Given....!}$

$\tt\large\green\rightarrow\bold{2x+Ky=1}$first equation

⠀

⠀

$\tt\large\green\rightarrow\bold{3x-5y=7}$seacond equation

$\tt\large\green\rightarrow\bold{2x+Ky=1........\times{3}}$

$\tt\large\green\rightarrow\bold{3x-5y=7}$

$\tt\large\green\rightarrow\bold{\cancel{6x}+3ky=3}$

⠀⠀$\tt\large\bold{\underline{{\cancel{6x}}-10y=14}}$

⠀⠀-⠀⠀⠀+⠀⠀⠀⠀⠀-

$\tt\large\green\rightarrow\bold{13Ky=11}$

$\tt\large\green\rightarrow\bold{y=\frac{11}{13k}}$

⠀

⠀

⠀

$\tt\large\bold{put.....{y=\frac{11}{13k}}\:in \;equation\: 1}$

$\tt\large\green\rightarrow\bold{2x+Ky=1}$

$\tt\large\green\rightarrow$2x + $\frac{11\cancel{k}} {13\cancel{k}}$ = 1

$\tt\large\green\rightarrow\bold{\frac{2x}{1}+ \frac{11}{13}}$=1

$\tt\large\green\rightarrow$$\frac{26x + 11}{13} = 1$

$\tt\large\green\rightarrow$$26x + 11 = 13$

$\tt\large\green\rightarrow$$\tt{26x = 13 + 11}$

$\tt\large\green\rightarrow$$26x = 24$

$\tt\large\green\rightarrow$$\tt{x = \frac{24}{26} }$

$\tt\large\green\rightarrow$$\tt{x = \frac{12}{13}}$

Equation 2$\huge\red{\ddot\smile}$

$\tt\large\green\rightarrow\bold{3x-5y=7}$

$\tt\large\green\rightarrow$$\tt{3 \frac{12}{13} - 5y = 7}$

$\tt\large\green\rightarrow$$\tt{\frac{36}{13} - \frac{5y}{1} = 7}$

$\tt\large\green\rightarrow$$\tt{\frac{36 - 65y}{13} = 7}$

$\tt\large\green\rightarrow$$\tt{36 - 65y = 91}$

$\tt\large\green\rightarrow$$\tt{- 65y = 91 - 36}$

$\tt\large\green\rightarrow$$\tt{- 65y = 55}$

$\tt\large\green\rightarrow$$\tt{y = \frac{55}{ - 65}}$

$\tt\large\green\rightarrow$$\tt\bold{y = \frac{11}{13} }$

$\tt\bold\blue{put \:the\: value \:of \:x \:and \:y\: in\: equation \:1..}$

⠀

$\tt\large\green\rightarrow\bold{2x+Ky=1}$

$\tt\large\green\rightarrow$$\tt\bold{2 (\frac{12}{13}) + k( \frac{11}{ - 13}) = 1}$

$\tt\large\green\rightarrow$$\tt{ \frac{24}{13} + \frac{11k}{ - 13} = 1}$

$\tt\large\green\rightarrow$$\tt{\frac{24 + - 11k}{13} = 1}$

$\tt\large\green\rightarrow\bold{24-11K=13}$

$\tt\large\green\rightarrow\bold{-11K = 13-24}$

$\tt\large\green\rightarrow\bold{-11K=-11}$

$\tt\large\green\rightarrow$$\tt{k = \frac{ - 11}{ - 11}}$

$\huge{\purple{\bigstar{\blue{\text{K=1 }}}}}$

$\huge\red{\ddot\smile}$ hope it helps you ✅✔️

Answer 2

Answer:

Given....!

\tt\large\green\rightarrow\bold{2x+Ky=1}→2x+Ky=1 first equation

⠀

⠀

\tt\large\green\rightarrow\bold{3x-5y=7}→3x−5y=7 seacond equation

\tt\large\green\rightarrow\bold{2x+Ky=1........\times{3}}→2x+Ky=1........×3

\tt\large\green\rightarrow\bold{3x-5y=7}→3x−5y=7

\tt\large\green\rightarrow\bold{\cancel{6x}+3ky=3}→

6x

+3ky=3

⠀⠀\tt\large\bold{\underline{{\cancel{6x}}-10y=14}}

6x

−10y=14

⠀⠀-⠀⠀⠀+⠀⠀⠀⠀⠀-

\tt\large\green\rightarrow\bold{13Ky=11}→13Ky=11

\tt\large\green\rightarrow\bold{y=\frac{11}{13k}}→y=

13k

11

⠀

⠀

⠀

\tt\large\bold{put.....{y=\frac{11}{13k}}\:in \;equation\: 1}put.....y=

13k

11

inequation1

\tt\large\green\rightarrow\bold{2x+Ky=1}→2x+Ky=1

\tt\large\green\rightarrow→ 2x + \frac{11\cancel{k}} {13\cancel{k}}

13

k

11

k

= 1

\tt\large\green\rightarrow\bold{\frac{2x}{1}+ \frac{11}{13}}→

1

2x

+

13

11

=1

\tt\large\green\rightarrow→ \frac{26x + 11}{13} = 1

13

26x+11

=1

\tt\large\green\rightarrow→ 26x + 11 = 1326x+11=13

\tt\large\green\rightarrow→ \tt{26x = 13 + 11}26x=13+11

\tt\large\green\rightarrow→ 26x = 2426x=24

\tt\large\green\rightarrow→ \tt{x = \frac{24}{26} }x=

26

24

\tt\large\green\rightarrow→ \tt{x = \frac{12}{13}}x=

13

12

Equation 2\huge\red{\ddot\smile}

⌣

¨

\tt\large\green\rightarrow\bold{3x-5y=7}→3x−5y=7

\tt\large\green\rightarrow→ \tt{3 \frac{12}{13} - 5y = 7}3

13

12

−5y=7

\tt\large\green\rightarrow→ \tt{\frac{36}{13} - \frac{5y}{1} = 7}

13

36

−

1

5y

=7

\tt\large\green\rightarrow→ \tt{\frac{36 - 65y}{13} = 7}

13

36−65y

=7

\tt\large\green\rightarrow→ \tt{36 - 65y = 91}36−65y=91

\tt\large\green\rightarrow→ \tt{- 65y = 91 - 36}−65y=91−36

\tt\large\green\rightarrow→ \tt{- 65y = 55}−65y=55

\tt\large\green\rightarrow→ \tt{y = \frac{55}{ - 65}}y=

−65

55

\tt\large\green\rightarrow→ \tt\bold{y = \frac{11}{13} }y=

13

11

\tt\bold\blue{put \:the\: value \:of \:x \:and \:y\: in\: equation \:1..}putthevalueofxandyinequation1..

⠀

\tt\large\green\rightarrow\bold{2x+Ky=1}→2x+Ky=1

\tt\large\green\rightarrow→ \tt\bold{2 (\frac{12}{13}) + k( \frac{11}{ - 13}) = 1}2(

13

12

)+k(

−13

11

)=1

\tt\large\green\rightarrow→ \tt{ \frac{24}{13} + \frac{11k}{ - 13} = 1}

13

24

+

−13

11k

=1

\tt\large\green\rightarrow→ \tt{\frac{24 + - 11k}{13} = 1}

13

24+−11k

=1

\tt\large\green\rightarrow\bold{24-11K=13}→24−11K=13

\tt\large\green\rightarrow\bold{-11K = 13-24}→−11K=13−24

\tt\large\green\rightarrow\bold{-11K=-11}→−11K=−11

\tt\large\green\rightarrow→ \tt{k = \frac{ - 11}{ - 11}}k=

−11

−11

\huge{\purple{\bigstar{\blue{\text{K=1 }}}}}★K=1