Question

if 3x+5y=9 and 5x+3y=7 then find x+y

Answer 1

3x +5y=9 5x+3y=7  now we multiple on both equation to make it same for easily eliminating. (3x+5y=9)*3 (5x+3y=7)*5   =6x+15y=27......(i)  =25x+15y=35.....(ii)   -       -        - Now we adding the both equations.  = -19x =  -8  X=8/19 X=1/2 ans.   Now we put the value of x in equation (i)   6x+15y =27 6*1/2+15y=27 3+15y=27 15y=27-3 15y=24 Y=24/15 ans.   Hope its helps u.    

Answer 2

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

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

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

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

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

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

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

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