Question

Please help with this question. Thank you!!!

Answer 1

`M=\left(-3, -(5)/(2)\right)`

`N=(3,-1)`

Explanation:

Given equations:

`4y=x-7`

`x^2+xy=4+2y^2`

To find the points of intersection, rearrange the linear equation to isolate x:

`\implies x=4y+7`

Substitute the found expression for x into the second equation:

`\implies (4y+7)^2+(4y+7)y=4+2y^2`

`\implies 16y^2+56y+49+4y^2+7y=4+2y^2`

`\implies 20y^2+63y+49=4+2y^2`

`\implies 18y^2+63y+45=0`

`\implies 9(2y^2+7y+5)=0`

`\implies 2y^2+7y+5=0`

Factor the quadratic and solve for y:

`\implies 2y^2+2y+5y+5=0`

`\implies 2y(y+1)+5(y+1)=0`

`\implies (2y+5)(y+1)=0`

`\implies 2y+5=0 \implies y=-(5)/(2)`

`\implies y+1=0 \implies y=-1`

Substitute the found values of y into the expression for x and solve for x:

`\begin{aligned}y=-(5)/(2) \implies x&=4\left(-(5)/(2)\right)+7\\x&=-10+7\\x&=-3\end{aligned}`

`\begin{aligned}y=-1 \implies x&=4\left(-1\right)+7\\x&=-4+7\\x&=3\end{aligned}`

Therefore, the coordinates of M and N are:

`M=\left(-3, -(5)/(2)\right)`

`N=(3,-1)`

Answer 2

M(-3, -2.5) and N(3, -1)

Explanation:

Given equations:

1st

• 4y = x - 7

make x subject

• x = 4y + 7 ___ equation 1

2nd

• x² + xy = 4 + 2y²

Substitute equation 1 into 2

x² + xy = 4 + 2y²

`\sf step : \ x = 4y + 7`

(4y + 7)² + (4y + 7)y = 4 + 2y²

`\sf step : \ distribute \ inside \ parenthesis`

16y² + 56y + 49 + 4y² + 7y = 4 + 2y²

`\sf step : \ collect \ terms`

16y² + 4y² -2y² + 56y + 7y + 49 - 4 = 0

`\sf step : \ simplify`

18y² + 63y + 45 = 0

`\sf step : \ middle \ term \ factor`

18y² + 18y + 45y + 45 = 0

`\sf step : \ factor \ out`

18y(y + 1) + 45(y + 1) = 0

`\sf step : \ collect \ into \ groups`

(18y + 45)(y + 1) = 0

`\sf step : \ set \ to \ zero`

y = -2.5, -1

Now, find value of x

x = 4y + 7

when y = -2.5, x = 4(-2.5) + 7 = -3

when y = -1, x = 4(-1) + 7 = 3

Hence the coordinates are (x, y) = M(-3, -2.5) and N(3, -1).