Question

Here is the 6th question with work, please

Answer 1

Hence, the solution of given system of equations is: x=-2, y=-4

Further explanation:

Given equations are:

$3x-2y=2\ \ \ \ Eqn\ 1\\5x-5y=10\ \ \ \ Eqn\ 2$

Multiplying eqn 1 by 5

$5(3x-2y)=5*2\\15x-10y=10\ \ \ \ \ Eqn\ 3$

Multiplying eqn 2 by 2

$2(5x-5y)=2*10\\10x-10y=20\ \ \ \ \ Eqn\ 4$

Subtracting Equation 4 from Equation 3

$15x-10y-(10x-10y)=10-20\\15x-10y-10x+10y = -10\\5x=-10\\\frac{5x}{5} = \frac{-10}{5}\\x=-2\\Putting\ x=2\ in\ eqn\ 1\\3(-2)-2y=2\\-6-2y=2\\-2y=2+6\\-2y=8\\\frac{-2y}{-2}=\frac{8}{-2}\\y=-4$

Proof:

Putting x=-2 and y=-4 in eqn 2

$5(-2)-5(-4)=10\\-10+20=10\\10=10$

Hence, the solution of given system of equations is: x=-2, y=-4

Keywords: Linear equations, Solution of linear system of equations

Learn more about linear equations at:

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