Question

Solve the following system of linear equations by cross multiplication method :
$2(ax-by)+a+4b=0$ and
$2(bx+ay)+b-4a=0$

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\underline{\textsf{Given,}} \\ \\ \sf \implies 2(ax \: - \: by) \: + \: a \: + \: 4b \: = \: 0 \\ \\ \sf \implies2ax \: - \: 2by \: + \: (a \: + \: 4b) \: = \: 0 \qquad...(1) \\ \\ \textsf{And,} \\ \\ \sf \implies 2(bx \: + \: ay) \: + \: b \: - \: 4a \: = \: 0 \\ \\ \sf \implies2bx \: + \: 2ay \: + \: (b \: - \: 4a) \: = \: 0 \qquad...(2)$





$\underline{\textsf{Using Cross Multiplication Method : }} \\ \\ \sf Coe. \: of \: x \qquad \: Coe .\: of \: y \qquad \: Const. term \\ \\ \sf \: \quad2a \qquad \qquad - 2b \qquad\qquad(a + \: 4b) \\ \\ \sf \: \quad2b \qquad \qquad \: \: \: \: \: 2a \qquad\qquad(b \: - \: 4a)$




$\sf \implies \dfrac{x}{-2b(b \: - \: 4a) \: - \: 2a(a \: + \: 4b)} \: = \: \dfrac{y}{2b(a \: + \: 4b) \: - \: 2a(b \: - \: 4a)} \: = \: \dfrac{1}{2a \: \cdot \: 2a \: - \: 2b \: \cdot \: (-2b)} \\ \\ \\ \sf \implies \dfrac{x}{ - 2 \{b(b \: - \: 4a) \: + \: a(a \: + \: 4b) \}} \: = \: \dfrac{y}{2 \{b(a \: + \: 4b) \: - \: a(b \: - \: 4a) \}} \: = \: \dfrac{1}{4 {a}^{2} \: + \: 4 {b}^{2} } \\ \\ \\ \sf \implies \dfrac{x}{ - 2 ( {b}^{2} \: - \: \cancel{4ab} \: + \: {a}^{2} \: + \: \cancel{ 4ab}) } \: = \: \dfrac{y}{2 ( \cancel{ab} \: + \: 4 {b}^{2} \: - \: \cancel{ab }\: + \: 4 {a}^{2}) } \: = \: \dfrac{1}{4 ({a}^{2} \: + \: {b}^{2} )}$




$\sf \implies \dfrac{x}{-2 \cancel{(a^2 \: + \: b^2)}} \: = \: \dfrac{y}{8 \cancel{(a^2 \: + \: b^2)}} \: = \: \dfrac{1}{4 \cancel{(a^2 \: + \: b^2)} } \\ \\ \\ \sf \implies \dfrac{ \: \: \: \: x \: \: \: \: }{ - 2} \: = \: \dfrac{y}{8} \: = \: \dfrac{1}{4} \\ \\ \\ \sf \implies \dfrac{x}{ - 1} \: = \: \dfrac{y}{4} \: = \: \dfrac{1}{2}$




$\underline{\textsf{On Comparison , }} \\ \\ \sf \implies \dfrac{x}{-1} \: = \: \dfrac{1}{2} \\ \\ \sf \: \therefore \: \: x \: = \: -\dfrac{1}{2} \\ \\ \textsf{And,} \\ \\ \sf \implies \dfrac{y}{4} \: = \: \dfrac{1}{2} \\ \\ \sf \implies y \: = \: \dfrac{4}{2} \\ \\ \sf\: \: \therefore \: \: y \: = \: 2$

Answer 2

Answer:

The given system of equation may be written as

2(ax - by) + a + 4b = 0

So, 2ax - 2by + a + 4b ..............(i)

2(bx + ay) + b - 4a = 0

so, 2bx+2ay+b-4a=0................(ii)

compare (i) and (ii) with standard form, we get

a1 = 2a, b1 = -2b, c1 = a + 4b

a2 = 2b, b2 = 2a, c2 = b - 4a

By cross multiplication method

x−2b2+8ab−2a2−8abx−2b2+8ab−2a2−8ab =−y2ab−8a2−2ab−8b2=−y2ab−8a2−2ab−8b2 =14a2+4b2=14a2+4b2

x−2b2−2a2=−y−8a2−8b2=14a2+4b2x−2b2−2a2=−y−8a2−8b2=14a2+4b2

Now, x−2b2−2a2=14a2+4b2x−2b2−2a2=14a2+4b2

⇒x=−12⇒x=−12

And, −y−8a2−8b2=14a2+4b2−y−8a2−8b2=14a2+4b2

⇒y=2⇒y=2

Therefore, the solution of the given pair of equations are −12−12 and 2 respectively.