Question

Can someone help me to solve a system of Equations Algebraically, please? I keep missing up :(​

Answer 1

Answer:

$x = - 1$

$y = - 3$

Step-by-step explanation:

$6x - 5y =9 \: \: \: \: \: \: \: \: ...(1)\\ 2x-y=1 \: \: \: \: \: \: \: \: \: \: ...(2)$

are given two equation

now,

from equation (1) we get,

$6x - 5y =9 \\ 6x = 5y + 9 \\ x = \frac{5y + 9}{6}$

now put the value of x in equation 2

$2( \frac{5y + 9}{6} ) - y = 1 \\ \\ \frac{10y + 18}{6} - y = 1 \\ \\ \frac{10y + 18 - 6y}{6} = 1 \\ \\ 10y + 18 - 6y = 6 \\ 4y = 6 - 18 \\ 4y = - 12 \\ y = - \frac{12}{4} \\ \\ y = - 3$

now put y = -3 in equation 2

$2x - ( - 3) = 1 \\ 2x + 3 = 1 \\ 2x = 1 - 3 \\ 2x = - 2 \\ x = - \frac{2}{2} \\ x = - 1$