Question

. Solve for x and y : 23x + 35y = 209 35x + 23y = 197

Answer 1

here is your answer that is 4,3

Answer 2

$\underline{\mathbf{Question}}$
Solve for x and y

23x + 35y = 209 ...(i)
35x + 23y = 197 ...(ii)

$\underline{\underline{\huge\mathfrak{Solution}}}$

$\underline{\mathbf{Adding \: both \: the \: equations \: we \: get}}$

$(23x + 35y ) + (35x + 23y) = 209 + 197 \\ \\ 23x + 35y + 35x + 23y = 406 \\ \\ 23x + 35x + 23y + 35y = 406 \\ \\ 58x + 58y = 406 \\ \\ \bf Can \: also \: be \: written \: as \\ \\ 58(x + y) = 58(7) \\ \\ \bf x + y = 7 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(iii)$

$\underline{\mathbf{Subtracting \: {eq}^{n} \: (ii) \: from \: {eq}^{n} \: (i) \: we \: get}}$

$(23x + 35y) - (35x + 23y) = 209 - 197 \\ \\ 23x + 35y - 35x - 23y = 12 \\ \\ - 12x + 12y = 12 \\ \\ \bf Can \: also \: be \: written \: as \\ \\ 12( - x + y) = 12(1) \\ \\ \bf - x + y = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(iv)$

Now,

Solving ${eq}^{n}$ (iii) and (iv) with elimination method :

$\underline{\mathbf{Subtracting \: {eq}^{n} \: (iv) \: from \: {eq}^{n} \: (iii)}}$

$(x + y) - ( - x + y) = 7 - 1 \\ \\ x + y + x - y = 6 \\ \\ x + x + y - y = 6 \\ \\ 2x = 2(3) \\ \\ \bf x = 3$

Putting the value of x in ${eq}^{n}$ (iii) we get,

$3 + y = 7 \\ \\ y = 7 - 3 \\ \\ \bf y = 4$

$\bf \therefore the \: value \: of \: x \: and \: y \: is \: 3 \: and \: 4.$