How do I solve he system 2x 1 3y 11 and 2x 3y 1 26 How

Solution

2^{(x + 1)} + 3^y = 11 ---------- (1) , 2^x - 3^{(y + 1)} = -26 ------ (2)

multiply equation (2) with 2

==> 2 * (2) ==> 2[2^x - 3^{(y + 1)}] = 2(-26)

==> 2^{(x +1)} - 2(3)^{(y +1)} = -52 ------- (3)       since a^m * aⁿ = a^m+n

(1) - (3) ==>

2^{(x + 1)} + 3^y - (2^{(x +1)} - 2(3)^{(y +1)} ) = 11 - (-52)

==> 2^{(x + 1)} + 3^y - 2^{(x +1)} + 2(3)^{(y +1)} = 63

==> 3^y + 2(3)^{(y +1)} = 63

==> 3^y + 2(3)(3)^y = 63         since a^m+n = a^m * aⁿ

==> 3^y + 6(3)^y = 63

==> 7(3^y) = 63

==> 3^y = 63/7 = 9

==> 3^y = 3²

==> y = 2

now substitute y = 2 in equation (1)

==> 2^{(x + 1)} + 3² = 11

==> 2^{(x + 1)} = 11 - 9

==> 2^{(x + 1)} = 2

==> 2^{(x + 1)} = 2¹

==> x +1 = 1

==> x = 0

Hence x = 0 , y = 2