Question

How do i determine the point of intersection of the linear functions 2x − y = 4 and x + 2y + 3 = 0?

Answer 1

`(1,-2)`

Explanation:

Solve them simultaneously; in this case elimination seems the better option.

`2x-y=4---(1)`

`x+2y=-3--(2)`

multiply `(1)` by `" "2" "` to get the coefficients of `y` the same.

`4x-2y=8--(1a)`

`x+2y=-3--(2)`

`(1a)+(2)`

`5x=5`

`=>x=1`

substitute into`" " (2)`

`1+2y=-3`

`2y=-4`

`=>y=-2`

check in #(1)

LHS

`2xx1- -2=2+2=4`

`=`RHS

intersection points`" "(1,-2)`

Answer 2

`(1,-2)`

Explanation:

`2x-color(red)(y)=4to(1)`

`x+2color(red)(y)+3=0to(2)`

`"from " (1) " we can express y in terms of x"`

`rArrcolor(red)(y)=2x-4to(3)`

`" substitute this value of y into " (2)`

`rArrx+2(2x-4)+3=0`

`rArrx+4x-8+3=0larr" distributing"`

`rArr5x-5=0larr" simplifying"`

`"add 5 to both sides"`

`5xcancel(-5)cancel(+5)=0+5`

`rArr5x=5rArrx=1`

`"substitute this value into " (3)" and evauate for y"`

`y=2-4=-2`

`color(blue)"As a check"`

Substitute these values into ( 2) and if equation is true the these are the solution.

`1+(2xx-2)+3=0rarr" True"`

`rArr"point of intersection "=(1,-2)`
graph{(y-2x+4)(y+1/2x+3/2)=0 [-10, 10, -5, 5]}