Question

(ii) ax + by=0
bx + ay=1+C
pls answer it step by step​

Answer 1

Answer:

kya haal hai partner

Step-by-step explanation:

According to the question,

we have, ax+by=c,bx+ay=1+c

suppose......

ax+by=c, -----------(1)

bx+ay=1+c------------(2)

(for finding the y co-ordinate),

multiply by b into equation-------------(1)

abx+b

2

y=bc------A

multiply by a into equation------------(2)

abx+a

2

y=a+ac-------B

then,subtracting into equ A &B,\begin{array}{l} \underline { \begin{array}{l} abx+{ b^{ 2 } }y=bc\, \\ { abx }_{ - }+_{ - }{ a^{ 2 } }y=a_{ - }+a_{ - }c \end{array} } \\ y({ b^{ 2 } }-{ a^{ 2 } })=bc-a-ac \\ y=\frac { { c\, ({ b^{ 2 } }-{ a^{ 2 } })-a } }{ { { b^{ 2 } }-{ a^{ 2 } } } } \end{array}

And, (for finding the x co-ordinate),

multiply by a into equation-------------(1)

a

2

x+aby=ac------------A

multiply by b into equation-------------(2)

b

2

x+aby=b+bc-------------B

then, subtracting into equ A & B,

\begin{array}{l} \underline { \begin{array}{l} { a^{ 2 } }x+aby=ac\, \\ { { b^{ 2 } }x }_{ - }+_{ - }aby=b_{ - }+b_{ - }c \end{array} } \\ x\, ({ a^{ 2 } }-{ b^{ 2 } })=ac-b-bc \\ x=\frac { { c\, (a-b)-b } }{ { { a^{ 2 } }-{ b^{ 2 } } } } \end{array}

So, we have x and y value.