Question

Solve the given equations

5x+2y+2=0,3x+4y-10=0

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Answer 1

Answer:

Given :

$\mapsto$ $\sf 5x + 2y + 2 =\: 0$

$\mapsto$ $\sf 3x + 4y - 10 =\: 0$

To Find :-

• What is the value of x and y.

Solution :-

Given equation :

$\leadsto \sf 5x + 2y =\: - 2\: -----\: (Equation\: No\: 1)$

$\leadsto \sf 3x + 4y =\: 10\: -----\: (Equation\: No\: 2)$

Now, from the equation no 1,

↦ $\sf 5x + 2y =\: -2$

↦ $\sf 5x =\: - 2 - 2y$

➦ $\sf\bold{\pink{x =\: \dfrac{- 2 - 2y}{5}}}$

Now, by putting the value of x in the equation no 2 we get,

⇒ $\sf 3\bigg(\dfrac{- 2 - 2y}{5}\bigg) + 4y =\: 10$

⇒ $\sf \dfrac{- 6 - 6y}{5} + 4y =\: 10$

⇒ $\sf - 6 - 6y + 20y =\: 5(10)$

⇒ $\sf - 6 + 14y =\: 50$

⇒ $\sf 14y =\: 50 + 6$

⇒ $\sf 14y =\: 56$

⇒ $\sf y =\: \dfrac{\cancel{56}}{\cancel{14}}$

➲ $\sf\bold{\red{y =\: 4}}$

Again, now by putting y = 4 in the equation no 1 we get,

⇒ $\sf 5x + 2(4) =\: - 2$

⇒ $\sf 5x + 8 =\: - 2$

⇒ $\sf 5x =\: - 2 - 8$

⇒ $\sf 5x =\: - 10$

⇒ $\sf x =\: \dfrac{- \cancel{10}}{\cancel{5}}$

➲ $\sf\bold{\red{x =\: - 2}}$

$\therefore$ The value of x is - 2 and the value of y is 4.

Answer 2

$\boxed{\begin{array}{c} \tt5x + 2y + 2 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt5x + 2y = \: \green{ - 2} \: \: \: \: \: \rm \cdots(i) \: \: \: \: \: \: {\Lleftarrow} \{\times 3 \}\end {array}} \\ \boxed{\begin{array}{c} \tt3x + 4y - 10 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt3x + 4y = \: \green{ 10} \: \: \: \: \: \rm \cdots(ii) \: \: \: \: \: \: {\Lleftarrow} \{\times 5 \}\end {array}}$

Equation (i) becomes

→ 15x + 6y = - 6 ----(a)

Equation (ii) becomes

→ 15x + 20y = 50 ----(b)

Subtracting Equation (a) from Equation (b), we get,

$\begin{array}{c} \: \: \: \: \: \: \: \: \: \rm15x + 20y = 50 \\ {\tiny{ \sf{minus}}}\: \: \: \underline{\rm15x + 6y \: \: = \: - 6} \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 14y = 56 \\ \\ \: \: \: \: \: \: \: \: \rightarrowtail \tt \dfrac{ \cancel{14}y}{\cancel{14}} = \dfrac{56}{14} \\ \\ \: \: \: \: \: \rightarrowtail \tt \: \: \: \: \: \: y = \red{4} \end {array}$

From Equation (i),

$\rm{}x = \frac{ - 2 - 2y}{5} \\ \\ \tt x = \frac{ - 2 - 2(4)}{5} = \frac{ - 2 - 8}{5} \\ \\ \tt x = \frac{ - 10}{5} \: \: \: \: = \red{ - 2}$

Thus,

• Value of x = -2
• Value of y = 4