Question

Find values of a and b which for simultanious equation x + 2y = 1 and ( a - b)x + ( a + b) y = a+b - 2


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Answer 1

Answer:

$\sf \huge \purple{Given :-}$

☣Two equation :

1. x + 2y = 1
2. (a - b)x + (a + b)y = (a + b) - 2

❖To Find :-

What is the value of a and b.

$‌\huge\mathcal{\fcolorbox{cyan}{black}{\pink{❥ᴀ᭄ɴsᴡᴇʀ࿐}}}$

Given two equation :

$\bold{\mapsto↦ \sf x + 2y = 1x+2y=1 \begin{gathered}\sf \\ \implies \bold{x + 2y - 1 =\: 0\: -----\: Equation\: no\: (1)}\\ \end{gathered} }$

$\sf\mapsto↦ \sf (a - b)x + (a + b)y = (a + b) - 2(a−b)x+(a+b)y=(a+b)−2$

$\begin{gathered}\sf \\ \implies \bold{(a - b)x + (a + b)y - (a + b + 2) =\: 0\: -----\: Equation\: no\: (2)}\\ \end{gathered} </p><p>$

Now, as we know that,

$\implies{\pink{\boxed{\large{\bold{\dfrac{a_1}{a_2} =\: \dfrac{b_1}{b_2} =\: \dfrac{c_1}{c_2}}}}}}$

From this formula we get,

$\sf a_1 =\: 1a$

$\sf b_1 =\: 2b$

$\sf c_1 =\: - 1c$

$\sf a_2 =\: (a - b)a$

$\sf b_2 =\: (a + b)b$

$\sf c_2 =\: - (a + b - 2)c$

Then we get,

$\implies\sf \purple{ \dfrac{1}{a - b} =\: \dfrac{2}{a + b} =\: \dfrac{- 1}{- (a + b - 2)} }$

Now, by taking first two parts we get,

$⇒ \sf \purple{ \dfrac{1}{a - b} =\: \dfrac{2}{a + b} }$

By doing cross multiplication we get,

$⇒ \sf \red{ 2(a - b) =\: a + b2(a−b)=a+b}$

$⇒ \sf \red{ 2a - 2b =\: a + b2a−2b=a+b}$

$⇒ \sf \red{2a - a =\: 2b + b2a−a=2b+b}$

$\begin{gathered}\sf \implies \bold{\pink{a =\: 3b -----\: Equation\: no\: (3)}}\\ \end{gathered}$

Again, by taking last two parts we get,

$↦ \sf \purple{\dfrac{2}{a + b} =\: \dfrac{{\cancel{-}} 1}{{\cancel{-}} (a + b - 2)} }$

By doing cross multiplication we get,

$↦ \sf \red{2(a + b - 2) =\: a + b2(a+b−2)=a+b}$

$↦ \sf \red{ 2a + 2b - 4 =\: a + b2a+2b−4=a+b}$

$↦ \sf \red{2a - a + 2b - b =\: 42a−a+2b−b=4}$

$\begin{gathered}\sf \implies \bold{\pink{a + b =\: 4\: -----\: Equation\: no\: (4)}}\\ \end{gathered}$

Now, by putting the value of equation no (3) in the equation no (4) we get,

$↛ \sf \red{3b + b =\: 43b+b=4}$

$↛ \sf \red{4b =\: 44b=4}$

$↛ \sf \red{b =\: \dfrac{\cancel{4}}{\cancel{4}}}$

$➠ \sf \huge\bold{\purple{b =\: 1}}$

Again, by putting the value of b in the equation no (3) we get,

$↛ \sf \red{a = 3ba=3b}$

$↛ \sf \red{a = 3(1)a=3(1)}$

$↛ \sf \red{a =\: 3 \times 1a=3×1}$

$➠ \sf \huge\mathfrak{\purple{a =\: 3}}$

$\sf \red{ The \: value \: of \: a \: is \: } {\huge \mathfrak\purple{ 3 }} \sf \red {\: and \: value of \: b \: is } \huge \mathfrak \purple{ \: 1 .}$

Answer 2

Explanation:

Answer:

$\sf \huge \purple{Given :-}$

☣Two equation :

x + 2y = 1

(a - b)x + (a + b)y = (a + b) - 2

❖To Find :

What is the value of a and b.

$‌\huge\mathcal{\fcolorbox{cyan}{black}{\pink{❥ᴀ᭄ɴsᴡᴇʀ࿐}}}‌$

Given two equation :

$\begin{gathered} \bold{\mapsto↦ \sf x + 2y = 1x+2y=1 \begin{gathered}\sf \\ \implies \bold{x + 2y - 1 =\: 0\: -----\: Equation\: no\: (1)}\\ \end{gathered} }\end{gathered}$

$\sf\mapsto↦ \sf (a - b)x + (a + b)y = (a + b) - 2(a−b)x+(a+b)y=(a+b)−2$

$\begin{gathered}\begin{gathered}\sf \\ \implies \bold{(a - b)x + (a + b)y - (a + b + 2) =\: 0\: -----\: Equation\: no\: (2)}\\ \end{gathered} < /p > < p > \end{gathered}$

</p><p>

Now, as we know that,

$\implies{\pink{\boxed{\large{\bold{\dfrac{a_1}{a_2} =\: \dfrac{b_1}{b_2} =\: \dfrac{c_1}{c_2}}}}}}$

From this formula we get,

$\sf a_1 =\: 1aa$

$\sf b_1 =\: 2bb$

$\sf c_1 =\: - 1cc$

$\sf a_2 =\: (a - b)$

$\sf b_2 =\: (a + b)$

$\sf c_2 =\: - (a + b - 2)$

Then we get,

$\implies\sf \purple{ \dfrac{1}{a - b} =\: \dfrac{2}{a + b} =\: \dfrac{- 1}{- (a + b - 2)} }$

Now, by taking first two parts we get,

$⇒ \sf \purple{ \dfrac{1}{a - b} =\: \dfrac{2}{a + b} }$

By doing cross multiplication we get,

$⇒ \sf \red{ 2(a - b) =\: a + b2(a−b)=a+b}$

$⇒ \sf \red{ 2a - 2b =\: a + b2a−2b=a+b}$

$⇒ \sf \red{2a - a =\: 2b + b2a−a=2b+b}$

$\begin{gathered}\begin{gathered}\sf \implies \bold{\pink{a =\: 3b -----\: Equation\: no\: (3)}}\\ \end{gathered} \end{gathered}$

Again, by taking last two parts we get,

$↦ \sf \purple{\dfrac{2}{a + b} =\: \dfrac{{\cancel{-}} 1}{{\cancel{-}} (a + b - 2)}}$

By doing cross multiplication we get,

$↦ \sf \red{2(a + b - 2) =\: a + b2(a+b−2)=a+b}$

$↦ \sf \red{ 2a + 2b - 4 =\: a + b2a+2b−4=a+b}$

$↦ \sf \red{2a - a + 2b - b =\: 42a−a+2b−b=4}$

$\begin{gathered}\begin{gathered}\sf \implies \bold{\pink{a + b =\: 4\: -----\: Equation\: no\: (4)}}\\ \end{gathered}\end{gathered}$

Now, by putting the value of equation no (3) in the equation no (4) we get,

$↛ \sf \red{3b + b =\: 43b+b=4}$

$↛ \sf \red{4b =\: 44b=4}↛4b=44b=4$

$↛ \sf \red{b =\: \dfrac{\cancel{4}}{\cancel{4}}}$

$➠ \sf \huge\bold{\purple{b =\: 1}}➠b=1$

Again, by putting the value of b in the equation no (3) we get,

$↛ \sf \red{a = 3ba=3b}$

$↛ \sf \red{a = 3(1)a=3(1)}$

$↛ \sf \red{a =\: 3 \times 1a=3×1}$

$➠ \sf \huge\mathfrak{\purple{a =\: 3}}➠a=3$

$\sf \red{ The \: value \: of \: a \: is \: } {\huge \mathfrak\purple{ 3 }} \sf \red {\: and \: value of \: b \: is } \huge \mathfrak \purple{ \: 1 .}$