Solve the following pairs of linear equations | vedclass

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(D) Given equations are:$8y - 3x = 5xy$  ---$(1)$$6y - 5x = -2xy$ ---$(2)$Case $1$: If $x = 0$,then from $(1)$,$8y = 0 \implies y = 0$. Checking in $(

Vedclass · Accepted Answer

$(0, 0), \left(\frac{22}{31}, \frac{11}{23}\right)$

Vedclass · Answer

(D) Given equations are:$8y - 3x = 5xy$  ---$(1)$$6y - 5x = -2xy$ ---$(2)$Case $1$: If $x = 0$,then from $(1)$,$8y = 0 \implies y = 0$. Checking in $(2)$,$6(0) - 5(0) = -2(0)(0) \implies 0 = 0$. Thus,$(0, 0)$ is a solution.Case $2$: If $x 
eq 0$ and $y 
eq 0$,divide both equations by $xy$:$(1)$ $\implies \frac{8}{x} - \frac{3}{y} = 5$$(2)$ $\implies \frac{6}{x} - \frac{5}{y} = -2$Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.$8u - 3v = 5$ ---$(3)$$6u - 5v = -2$ ---$(4)$Multiply $(3)$ by $5$ and $(4)$ by $3$:$40u - 15v = 25$$18u - 15v = -6$Subtracting: $22u = 31 \implies u = \frac{31}{22} \implies x = \frac{22}{31}$.Substitute $u$ in $(3)$: $8(\frac{31}{22}) - 3v = 5 \implies \frac{124}{11} - 3v = 5 \implies 3v = \frac{124}{11} - 5 = \frac{69}{11} \implies v = \frac{23}{11} \implies y = \frac{11}{23}$.Thus,the solutions are $(0, 0)$ and $\left(\frac{22}{31}, \frac{11}{23}ight)$.

Part $I$	Part $II$
$1. x+2y=3, 2x+4y=5$	$a. \text{Solution set is a singleton set.}$
$2. 2x+3y=6, x+\frac{3}{2}y=3$	$b. \text{Solution set is an empty set.}$
$3. 3x-y=0, x-y+6=0$	$c. \text{Solution set is an infinite set.}$
	$d. \text{Solution set contains two members.}$

Solve the following pairs of linear equations: $8y - 3x = 5xy$,$6y - 5x = -2xy$.

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