Question

Please help, thank you!!

Answer 1

Coordinates of M and N are

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

The order is not important since we are not told where M and N points actually are. The attached graph will make this clear

Explanation:

The two equations are

`4y=x-7 \textrm{\space ( Eq 1)}`

and

`x^2 + xy = 4 + 2y^2 \textrm{ (Eq 2)}`

Adding 7 to both sides of Eq 1 gives us

`x = 4y + 7`

Eq2 can be re-written as

`x^2 + xy - 4 - 2y^2 = 0 \textrm{ (Eq 3)}`

The above is done by subtracting
`4 + 2y^2` on both sides of Eq 2

Substituting for
`x = 4y + 7` in Eq 3 gives us

`\left(4y+7\right)^2+\left(4y+7\right)y-4-2y^2 = 0`

Let's examine the first two terms and simplify

`(4y+7)^2 = \left(4y\right)^2+2\cdot \:4y\cdot \:7+7^2` =
`16y^2+56y+49` (A)

`\left(4y+7\right)y = 4y^2 + 7y` (B)

So Eq 3 can be reduced to a quadratic equation with one variable,
`y`

(A) + (B) -
`4 -2y^2` =
`6y^2+56y+49+4y^2+7y-4-2y^2`

Grouping like terms we get

`16y^2+4y^2-2y^2+56y+7y+49-4`

Adding similar terms we get

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

We can divide both sides of the equation by 9 to make it easier to solve using the equation for the solutions of a quadratic equation

Dividing both sides by 9 we get

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

Use the quadratic formula

`y_1, y_2 = ( -b \pm √(b^2 - 4ac))/( 2a )`

where a is the coefficient of
`y^2`,
`b` the coefficient of
`y` and
`c` the constant

Here
`a = 2, b = 7` and
`c = 5`

So we get

`y_1, y_2 = (-7\pm √(7^2-4\cdot \:2\cdot \:5))/(2\cdot \:2)` =
`(-7\pm \:3)/(2\cdot \:2)`

The above is obtained by noting that

`√(7^2-4\cdot \:2\cdot \:5)} = √(49-40) =√(9) = 3`

`y_1=(-7+3)/(2\cdot \:2),\:y_2=(-7-3)/(2\cdot \:2)`

`y_1 = (-7+3)/(2\cdot \:2) = (-4)/(4) = -1` (using the + component)

`y_2 = (-7-3)/(2\cdot \:2) = (-10)/(4) = -2.5`

Find
`x_1` and
`x_2` by substituting each of these values in the equation for
`x`

`x_1 = 4(-1) + 7 = -4 + 7 = 3`

Therefore the coordinates of M and N are

`(-3, -2.5)` and
`(3, -1)`

The order is not important since we are not told which are M and N points. See the attached graph for a visual depiction

Hope that helps :)

Answer 2

`\sf M =\left(-3, -(5)/(2)\right), \quad N= \left(3, -1 \right)`

Explanation:

Given:

`\begin{cases}4y = x - 7\\x^2+xy=4+2y^2\end{cases}`

To find the points at which the line and the curve intersect, rewrite the first equation to make y the subject by dividing both sides by 4:

`\implies y=(1)/(4)x-(7)/(4)`

Substitute the expression for y into the second equation and solve for x:

`\implies x^2+x\left((1)/(4)x-(7)/(4)\right)=4+2\left((1)/(4)x-(7)/(4)\right)^2`

`\implies x^2+(1)/(4)x^2-(7)/(4)x=4+2\left((1)/(16)x^2-(7)/(8)x+(49)/(16)\right)`

`\implies (5)/(4)x^2-(7)/(4)x=4+(1)/(8)x^2-(7)/(4)x+(49)/(8)`

`\implies (5)/(4)x^2-(7)/(4)x=(1)/(8)x^2-(7)/(4)x+(81)/(8)`

`\implies (5)/(4)x^2=(1)/(8)x^2+(81)/(8)`

Multiply both sides by 8:

`\implies 10x^2=x^2+81`

`\implies 9x^2=81`

`\implies x^2=9`

`\implies x=√(9)`

`\implies x=\pm3`

To find the y-values of the coordinates of the points of intersection (M and N), substitute the found values of x into the expression for y:

`x=3 \implies y=(1)/(4)(3)-(7)/(4)=-1`

`x=-3 \implies y=(1)/(4)(-3)-(7)/(4)=-(5)/(2)`

Therefore:

`\sf M =\left(-3, -(5)/(2)\right), \quad N= \left(3, -1 \right)`