Math, asked by afzalnishawahid13, 11 months ago

if 3
if \: 3x + 5y = 9 \:  \: and \: 5x + 3y = 7
then what is the value of
x + y

Answers

Answered by tanya345662
2

Answer:

3x+5y=9-----(1)

5x+3y=7-----(2)

multiplying eq(1) by 5 & eq(2) by 3 we get

15x+25y=45

-15x ± 9y= - 21

_____________

  16y=24

y = 24/16 = 3/2

and

x = 9-5x3/2/3 = -1/2

hence x = 1/2 and y = 3/2

so x + y = 1/2 + 3/2 = 2

Answered by TRISHNADEVI
7

\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}

\underline{ \mathfrak{ \: \: Given :- \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \text{ 3x + 5y = 9 \: \: \: - - - - - - > (1) } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \text{5x + 3y = 7 \: \: \: - - - - - - >(2) \: } \\ \\ \underline{ \mathfrak{ \: \: To \: \: find :- \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \text{ \huge{ x + y = ?}}

\underline{ \mathfrak { \:Now,\:}} \\ \\ \rm{(1) \times 3 \implies 9x + 15y = 27 \: \: - - - - > (3)} \\ \\ \rm{(2) \times 5 \implies 25x + 15y = 35 \: \: - - - - > (4)}

 \sf{(4) - (3) \implies \: 16x = 8} \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \sf{ \implies \: x =  \frac{8}{16} } \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \sf{ \therefore \: \red{x = \frac{1}{2}} }

\underline{ \text{ \: Putting the value of \: \red{x} \: in eq. (1) , we get, \: }}

\: \: \: \: \: \: \: \: \: \sf{ 3 \times ( \red{\frac{1}{2}} ) + 5y = 9 }\\ \\ \sf{\implies \: \frac{3}{2} + 5y = 9 }\\ \\ \sf{\implies \: 5y = 9 - \frac{3}{2}} \\ \\ \sf{ \implies \: 5y = \frac{18 - 3}{2}} \\ \\ \sf{\implies \: 5y = \frac{15}{2}} \\ \\ \sf{ \implies \: y = \frac{15}{2} \times \frac{1}{5} } \\ \\ \: \: \: \: \: \: \: \: \: \: \sf{\therefore \: \red{ y = \frac{3}{2} }}

\tt{ \therefore \: \pink{ x + y = \frac{1}{2} + \frac{3}{2} }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ \pink{ = \frac{1 + 3}{2} }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ \pink{ = \frac{4}{2} }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ \pink{ =2 }}

\: \: \: \: \: \: \: \:\: \: \huge{ \therefore \: \boxed{ \bold{ \green{ \: x + y = 2 \: }}}}

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