Question

Solve it by using cross multiplication method

${a}^{2} x + {b}^{2} y - {c}^{2} = 0\\ {b}^{2} x + {a}^{2}y - {d}^{2} = 0$
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Answer 1

Given systems of equations are:

2x−3y+4=0    ....(1)

2x−35y+12=0       ...(2)

Now from eq1 and eq2 will get:

3x−2y+24          ...(3)

3x−10y+72          ...(4)

To solve this pair of equations for x and y using cross-multiplication, we arrange the variables x and y and their coefficients a1, a2, b1 and b2, and the constants c1 and c2 as:

x=a1b2−a2b1b1c2−b2c1

y=a1b2−a2b1c1a2−c2a1

⇒x=(3×10)−(3×−2)(−2×72)−(−10×24)

⇒y=(3×10)−(3×−2)(24×3)−(72×3)

⇒x=−30−(−6)−144−(240)

⇒x=−2496=−4

⇒y=−30−(−6)72−216

⇒y=−24−144=6

⇒x=−4,y=6

x+y= 6−4=2

Answer 2

Answer:

$\mathtt\purple{LET:-}$

$\mathtt{a1 = {a}^{2} \: \: \: \: \: \: \: \: b1 = {b}^{2} \: \: \: \: \: \: \: c1 = { - c}^{2} }$

$\mathtt{a2 = {b}^{2} \: \: \: \: \: \: \: \: b2 = {a}^{2} \: \: \: \: \: \: \: c2 = { - d}^{2} }$

$\mathtt\purple{SOLUTION:-}$

$\mathtt\purple{USING \: FORMULA}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\mathtt{ \frac{x}{ {b}^{2} {( - d)}^{2} - {a}^{2} {( - c)}^{2} } = \frac{y}{ {( - c) {b}^{2} - ({d}^{2}) ({a}^{2}) {{} } } } = \frac{1}{ {a}^{2} {a}^{2} - {b}^{2} {b}^{2} } }$

$\mathtt{ \frac{x}{ - {b}^{2} {d}^{2} + {a}^{2} {c}^{2} } = \frac{y}{ - {b}^{2} {c}^{2} + {a}^{2} {d}^{2} } = \frac{1}{ {a}^{4} {b}^{4} } }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\mathtt \purple{x = \frac{ { - b}^{2} {a}^{2} + {a}^{2} {c}^{2} }{ {a}^{4} - {b}^{4} } \: \: \: \: \: \: \: \: \: y = \frac{ { - b}^{2} {c}^{2} + {a}^{2} {d}^{2} }{ {a}^{4} - {b}^{4} } }$

$\mathtt \purple{ x= \frac{ {a}^{2} {c}^{2} - {b}^{2} {a}^{2} }{ {a}^{4} - {b}^{4} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y= \frac{ {a}^{2} {d}^{2} - {b}^{2} {c}^{2} }{ {a}^{4} - {b}^{4} } }$