Question

Question #063b3

Answer 1

Assuming you are to find the intersection point or common solution to the equations.

`x=1`, `y=4`

Explanation:

`4x+2y=12`------eqn.1
`3x+y=7`---------eqn.2

Rearranging equation 2
`3x+y=7`
`y=7-3x`--------eqn.3

Substituting equation 3 into equation 1
`4x+2y=12`
`4x+2(7-3x)=12`

`4x+14-6x=12`
`2=2x`
`x=1`

Substituting `x=1` into any equation above to find the value of `y`

I'll use equation 3. (It is the simplest as it is a equation of `y` in terms of `x`

`y=7-3x`
`y=7-3`
`y=4`

`:.` `x=1`, `y=4`

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There is other approach to solve this question too,
eg. using matrices.

First arrange in a form that all like terms of both equation are at the same position. (The equations come with the form, that's good)

`4x+2y=12`
`3x+y=7`

`[(4,2),(3,1)][(x),(y)]=[(12),(7)]`

`[(x),(y)]=[(4,2),(3,1)]^-1[(12),(7)]`

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`[(4,2),(3,1)]^-1` is the inverse of `[(4,2),(3,1)]`

To find the inverse of a matrice`[(a,b),(c,d)]`, the inverse would be:
`1/(ad-bc)[(d,-b),(-c,a)]`

`:.` `[(4,2),(3,1)]^-1=1/(4-6)[(1,-2),(-3,4)]`
`=-1/2[(1,-2),(-3,4)]`
`=[(-1/2,1),(3/2,-2)]`

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`[(x),(y)]=[(4,2),(3,1)]^-1[(12),(7)]=[(-1/2,1),(3/2,-2)][(12),(7)]`
carrying out matrice multiplication

`[(x),(y)]=[(-6+7),(18-14)]`
`[(x),(y)]=[(1),(4)]`

`:.x=1, y=4`