Question

Solve the equation
3x +2y+2= 0.5x - y -27=0​

Answer 1

Question :-

Solve the equation

3x +2y+2= 0.5x - y -27=0

Solution :-

Given :

• 3x + 2y + 2 = 0
• 0.5x - y - 27 = 0

Need to find :

• x and y

By first,

$\sf\implies{3x + 2y + 2 = 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\sf\implies{3x + 2y = - 2} - - - (1)$

By second,

$\sf\implies{}0.5x - y - 27 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies{}0.5x - y = 27 - - - (2)$

Multiplying equation (2) by 2

$\sf\implies{}2(0.5x - y = 27) \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\sf\implies{1x - 2y = 54} - - - (3)$

Now, we have,

$\red{\sf:\longmapsto{We \: have \begin{cases} \sf3x + 2y = -2 \: - - - (1)\\ \sf x - 2y = 54 \: \: \: \: - - - (3) \end{cases}}}$

Adding both equation,

$\sf:\longmapsto{3x + 2y + (x - 2y) = - 2 + 54} \\ \\ \sf:\longmapsto{3x + x + 2y - 2y = 52} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{4x = 52} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{}x = \frac{52}{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{x = 13} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Substitute x = 13 in equation (3)

$\sf:\longmapsto{}13 - 2y = 54 \: \: \: \: \\ \\ \sf:\longmapsto{ - 2y = 54 - 1}3 \\ \\ \bf:\longmapsto{} \red{y = \frac{41}{ - 2} } \: \: \: \: \: \: \: \: \: \:$

$\red{\sf:\longmapsto{ \begin{cases} \sf{x = 13}\\ \\ \sf {y = \large{\frac{41}{-2}}}\end{cases}}}$

Verification :-

Substituting x and y in equation (3)

$\sf : \longmapsto{13 \: \cancel{- 2} \times \frac{41} { \cancel{- 2}} = 54} \\ \\ \sf : \longmapsto{13 + 41= 54} \: \: \: \: \: \: \: \: \\ \\ \bf : \longmapsto{} \red{54 = 54 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence Verified ✔

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Answer 2

Question :- Solve the equation.

$\sf{3x + 2y + 2 = 0.5x - y - 27 = 0}$

Given :-

• $\sf{→3x + 2y + 2 = 0}$
• $\sf{→0.5x - y - 27 = 0}$

To find :-

• x and y

Rearranging the ❶ equation,

$\sf{→3x + 2y + 2 = 0}$

$\sf {\therefore 3x + 2y = - 2}$-----❶

Rearranging the ❷ equation,

$\sf{→0.5x - y - 27 = 0}$

$\sf{ \therefore 0.5x - y = 27}$-----❷

Multiplying equation ❷ by 2,

$\sf{→2(0.5x - y = 27)}$

$\sf{→(1x - 2y = 54)}$-----❸

We now have,

$\sf {→3x + 2y = - 2}$-----❶

$\sf{→x - 2y = 54}$-----❸

By elimination method adding the equation ❶ and equation ❸,

$\sf {3x + 2y = - 2 }\\ \sf{ \: \: x - 2y = \: 54} \\------- \\ \sf{4x \: \: \: \: \: \: \: \: \: \: = \: 52} \\ -------$

We get,

$\sf{→x = \frac{52}{4} }$

$\sf \fbox{→x = 13}$

Now, substituting the value of x = 13 in eq ❸,

$\sf{→13 - 2y = 54}$

$\sf{→ - 2y = 54 - 13}$

$\sf{→ - 2y = 41}$

$\sf {→y = \frac{41}{ - 2}}$

Hence $\sf {x = 13}$ and $\sf {y = \frac{41}{ - 2}}$.

Verification :-

$\sf{→13 - 2( \frac{41}{ - 2})= 54}$

$\sf{→13 + 41= 54}$

$\sf{→54 = 54}$

LHS = RHS.

Hence Verified.

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