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rvkacademic

Answer:

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

Step-by-step explanation:

The two equations are

[tex]4y=x-7 \textrm{\space ( Eq 1)}[/tex]

and

[tex]x^2 + xy = 4 + 2y^2 \textrm{ (Eq 2)}[/tex]

Adding 7 to both sides of Eq 1 gives us

[tex]x = 4y + 7[/tex]

Eq2 can be re-written as

    [tex]x^2 + xy - 4 - 2y^2 = 0 \textrm{ (Eq 3)}[/tex]

The above is done by subtracting [tex]4 + 2y^2[/tex] on both sides of Eq 2

Substituting for [tex]x = 4y + 7[/tex] in Eq 3 gives us  

[tex]\left(4y+7\right)^2+\left(4y+7\right)y-4-2y^2 = 0[/tex]    

Let's examine the first two terms and simplify


[tex](4y+7)^2 = \left(4y\right)^2+2\cdot \:4y\cdot \:7+7^2[/tex] = [tex]16y^2+56y+49[/tex]   (A)

[tex]\left(4y+7\right)y = 4y^2 + 7y[/tex]      (B)

So Eq 3 can be reduced to a quadratic equation with one variable, [tex]y[/tex]

(A) + (B) - [tex]4 -2y^2[/tex] = [tex]6y^2+56y+49+4y^2+7y-4-2y^2[/tex]

Grouping like terms we get

[tex]16y^2+4y^2-2y^2+56y+7y+49-4[/tex]

Adding similar terms we get

[tex]18y^2+63y+45 = 0[/tex]

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

[tex]2y^2+7y+5 = 0[/tex]

Use the quadratic formula

[tex]y_1, y_2 = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }[/tex]

where a is the coefficient of [tex]y^2[/tex], [tex]b[/tex] the coefficient of [tex]y[/tex] and [tex]c[/tex] the constant

Here [tex]a = 2, b = 7[/tex] and [tex]c = 5[/tex]

So we get

[tex]y_1, y_2 = \frac{-7\pm \sqrt{7^2-4\cdot \:2\cdot \:5}}{2\cdot \:2}[/tex] = [tex]\frac{-7\pm \:3}{2\cdot \:2}[/tex]    

The above is obtained by noting that

[tex]\sqrt{7^2-4\cdot \:2\cdot \:5}} = \sqrt{49-40} =\sqrt{9} = 3[/tex]

[tex]y_1=\frac{-7+3}{2\cdot \:2},\:y_2=\frac{-7-3}{2\cdot \:2}[/tex]

[tex]y_1 = \frac{-7+3}{2\cdot \:2} = \frac{-4}{4} = -1[/tex]   (using the + component)

[tex]y_2 = \frac{-7-3}{2\cdot \:2} = \frac{-10}{4} = -2.5[/tex]

Find [tex]x_1[/tex] and [tex]x_2[/tex] by substituting each of these values in the equation for [tex]x[/tex]

[tex]x_1 = 4(-1) + 7 = -4 + 7 = 3[/tex]

Therefore the coordinates of M and N are

 [tex](-3, -2.5)[/tex] and [tex](3, -1)[/tex]

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 :)

Ver imagen rvkacademic
semsee45

Answer:

[tex]\sf M =\left(-3, -\dfrac{5}{2}\right), \quad N= \left(3, -1 \right)[/tex]

Step-by-step explanation:

Given:

[tex]\begin{cases}4y = x - 7\\x^2+xy=4+2y^2\end{cases}[/tex]

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:

[tex]\implies y=\dfrac{1}{4}x-\dfrac{7}{4}[/tex]

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

[tex]\implies x^2+x\left(\dfrac{1}{4}x-\dfrac{7}{4}\right)=4+2\left(\dfrac{1}{4}x-\dfrac{7}{4}\right)^2[/tex]

[tex]\implies x^2+\dfrac{1}{4}x^2-\dfrac{7}{4}x=4+2\left(\dfrac{1}{16}x^2-\dfrac{7}{8}x+\dfrac{49}{16}\right)[/tex]

[tex]\implies \dfrac{5}{4}x^2-\dfrac{7}{4}x=4+\dfrac{1}{8}x^2-\dfrac{7}{4}x+\dfrac{49}{8}[/tex]

[tex]\implies \dfrac{5}{4}x^2-\dfrac{7}{4}x=\dfrac{1}{8}x^2-\dfrac{7}{4}x+\dfrac{81}{8}[/tex]

[tex]\implies \dfrac{5}{4}x^2=\dfrac{1}{8}x^2+\dfrac{81}{8}[/tex]

Multiply both sides by 8:

[tex]\implies 10x^2=x^2+81[/tex]

[tex]\implies 9x^2=81[/tex]

[tex]\implies x^2=9[/tex]

[tex]\implies x=\sqrt{9}[/tex]

[tex]\implies x=\pm3[/tex]

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:

[tex]x=3 \implies y=\dfrac{1}{4}(3)-\dfrac{7}{4}=-1[/tex]

[tex]x=-3 \implies y=\dfrac{1}{4}(-3)-\dfrac{7}{4}=-\dfrac{5}{2}[/tex]

Therefore:

[tex]\sf M =\left(-3, -\dfrac{5}{2}\right), \quad N= \left(3, -1 \right)[/tex]