plz solve this simultaneous equation?

Question

the straight line 2x+y=2 meets the curve x^2+xy+3=0 at two distinct points .find the coodinates of these two points.

show workings

Answer 1

Do your own homework.

Answer 2

if you solve the linear equation for either x or y then you can plug that into the other equation and solve for the other variable. For example, lets solve the linear equation for y.

2x + y = 2
y = -2x + 2 (Subtracting 2x from both sides)

So now take this equation and plug it into the other equation.
Since we know that y is equal to -2x + 2, then the other equation is solved in this format...

x^2 + xy + 3 = 0
x^2 + x(-2x + 2) + 3 = 0 (Replacing y with -2x + 2)
x^2 -2x^2 + 2x + 3 = 0 (Using the distributive property)
-x^2 + 2x + 3 =0 (Combing like terms)
x^2 - 2x - 3 = 0 (Dividing thru by negative one so this equ can be factored)
(x - 3) (x + 1) = 0 (Factors of the quadratic equation above)
x = 3, x = -1 (Setting both factors equal to 0 and solving for x)

Replace the first x value which is 3 back into the linear equation and solve for y. So...

y = -2x + 2
y = -2(3) + 2
y = -4

So the first coordinate is (3, -4)

Replace the second x value which is -1 back into the linear equation and solve for y. So...

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

So the second coordinate is (-1, 4)

When you check both of these coordinates in the other equation, it checks out. Look below...

x^2 + xy + 3 = 0
3^2 + 3(-4) + 3 = 0
9 - 12 + 3 = 0
-3 + 3 = 0
0 = 0

x^2 + xy + 3 = 0
(-1)^2 + -1(4) + 3 = 0
1 - 4 + 3 = 0
-3 + 3 = 0
0 = 0

I hope that this helps you out!!!!

Answer 3

you have the equation
2X+Y=2 if you solve for Y you will get Y=2-2X
you can put the value of Y in equation 2
like this:
X^2+X(2-2X)+3=0
X^2+2X-2X^2+3=0
-X^2+2X+3=0
X^2-2X-3=0 (we flip the signs)
(x-3)(x+1) (we factor)
x=3 and x=-1

now you can put this values of x in the equation Y=2-2X
Y=2-2(3) equal Y=-4
y=2-2(-1) equals Y=4

so the two points of intersection are (3,-4) and (-1,-4)

Answer 4

2x+y=2 means y=2(1-x). Substitute for y in the other equation:
x^2+2x(1-x)+3=0
x^2+2x-2x^2+3=0
-x^2+2x+3=0
x^2-2x-3=0
x^2+x-3x-3=0 (note that x-3x is the same as -2x)
x(x+1)-3(x+1)=0 (x+1 is common, so factor it out)
(x-3)(x+1)=0, meaning x=3 or x=-1.
Now, when x=3, y=-4;
when x=-1, y=4.

Answer 5

2x + y = 2
x^2 + xy + 3 = 0

2x + y = 2
y = -2x + 2

x^2 + (-2x + 2)x + 3 = 0
x^2 - 2x^2 + 2x + 3 = 0
(1 - 2)x^2 + 2x + 3 = 0
-x^2 + 2x + 3 = 0
-(x^2 - 2x - 3) = 0
-(x - 3)(x + 1) = 0
x = 3 or -1

y = -2x + 2
y = -2(3) + 2
y = -6 + 2
y = -4

y = -2(-1) + 2
y = 2 + 2
y = 4

ANS : (3,-4) and (-1,4)

Answer 6

solve each equation for y:
y=2 - 2x and y = -(3+x^2)/x
so solving (2 - 2x)=-(3+x^2)/x for x will give you you answer:
=> 2x - 2x^2= -3 -x^2
=> x^2 - 2x - 3 = 0
=> (x+1)(x-3)=0
=> x1=-1 , x2=3
=> P1(-1,4), P2(3,-4)

Answer 7

make y the subject... so it's y=2-2x.... substitute y=2-2x into curve equation...so it will be x^2+x(2-2x)+3=0... then u solve for x.... then solve for y...

Answer 8

Isn't that what your teacher asked you to do?

Answer 9

(3,-4)
(-1,4)