Question

Solve the system of equations using the elimination method 4x+5y=40 6x+3y=42

Answer 1

The solutions to the system of equations are
`y=4,\:x=5`.

Explanation:

To solve the system
`\begin{bmatrix}4x+5y=40\\ 6x+3y=42\end{bmatrix}`

First,

`\mathrm{Multiply\:}4x+5y=40\mathrm{\:by\:}3\:\mathrm{:}\:\quad \:12x+15y=120\\\\\mathrm{Multiply\:}6x+3y=42\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:12x+6y=84`

`\begin{bmatrix}12x+15y=120\\ 12x+6y=84\end{bmatrix}`

Subtract the first equation from the second equation

`12x+6y=84\\\underline{-12x-15y=-120}\\-9y=-36`

Solve
`-9y=-36` for y:

`(-9y)/(-9)=(-36)/(-9)\\y=4`

For
`12x+15y=12` plug in
`y=4` and solve for x

`12x+15\cdot \:4=120\\12x=60\\x=5`

The solutions to the system of equations are:

`y=4,\:x=5`