Solve the following systems equations?

Question

x*x-2y=11
x=y+4

show working please

Answer 1

(y+4)(y+4)-2y -11=0
y^2 +8y + 16 -2y -11 =0
y^2 +6y +5=0
(y+5)(y+1)=0
y=-5, x=-1 , and y=-1, x=3

Check with y=-5
(-1)^2 -2(-5) = 1+10=11 OK

Check with y=-1
3^2 -2(-1) = 9+2=11 OK

Answer 2

These are two simultaneous equation but the first one has a x*x or x^2 term and is a quadratic. There will be two solutions This is how to do it:

x = y + 4

==> y = x - 4

Now substitute x - 4 for y in the other equation.

==> x*x - 2(x - 4) =11

==> x^2 - 2x + 8 = 11

==> x^2 - 2x - 3 = 0 This is the familiar quadratic

Factorising:
==> x^2 -3x + x -3 = 0

==> x(x - 3) + 1(x - 3) =0

==> (x - 3) (x + 1) = 0

==> (x - 3) = 0 or (x + 1) = 0

==> x = 3 or x = -1

Substitute each of these values for x in turn to get the corresponding values of y.

x = 3

==> 3 = y + 4

==> y = -1 So 1st solution is: x = 3 and y = -1

When x = -1

==> -1 = y + 4

==> y = -5 So 2nd solution is x = -1 and y = -5

Always check.

When x = 3 and y = -1

==> 3*3 - 2*(-1) = 11
==> 9 + 2 = 11 which is correct

When x = -1 and y = -5

==> -1*(-1) - 2(-5) = 11
==> 1 + 10 = 11 which is correct

Answer 3

Assuming x*x means x squared = x^2
Substitute the y+4 into the first equation for x ==>
(y+4)*(y+4) -2y = 11 expand the product ==>
y^2 +8y +16 -2y = 11 combine like terms and solve for 0 ==>
y^2 +6y + 5 = 0 factor this equation to
(y+5)(y+1) = 0 set each factor = 0 and solve
y+5=0 ==> y = -5
y+1=0 ==> y = -1 substituting these into the first equation ==>
x*x - 2(-5) = 11 ==> x^2 + 10 = 11 ==> x^2 = 1 ==> x=+1 or -1
x*x - 2(-1) = 11 ==> x^2 + 2 = 11 ==> x^2 = 9 ==> x=+3 or -3
The possible answers are:
x = 1 and y = -1
x = 1 and y = -5
x = -1 and y = -1
x = -1 and y = -5
x = 3 and y = -1
x = 3 and y = -5
x = -3 and y = -1
x = -3 and y = -5
Substituting back into the original equations you will find that only
x = 3 and y = -1 or
x = -1 and y = -5 solve both equations.

Answer 4

You have x*x-2y=11 and x=y+4
Substitute x=y+4 in x*x-2y=11
(y+4)*(y+4) -2y = 11
=> y^2 +4y +4y +16 -2y = 11
=> y^2 +6y +16 -11 = 0
=> y^2 +6y + 5 = 0
=> y^2 +y + 5y + 5 = 0
=> y(y +1 ) + 5(y + 1) = 0
=> (y +1 )(y + 5) = 0
=> (y +1 )=0 or (y + 5) = 0
=> y = -1 or y = -5

Now, x=y+4,
When y = -1 , x = (-1)+4 = 3
And when y = -5 ,x = (-5)+4 = -1

So, your system has two solutions
(x,y) = (3 , -1) and (x,y)=(-1,-5)

Answer 5

If there's a quadratic, then there'll likely be 2 solutions. first of all, the addition technique could paintings completely in this technique. upload the main astounding factors and the left factors to get. y + 5x + 6x^2 - 13x - y = 3 - 5 word how the 'y's cancel out 6x^2 - 8x = -2 Divide the two factors by utilising 2 3x^2 - 4x = -a million 3x^2 - 4x + a million = 0 (3x - a million)(x-a million) = 0 x = a million ; a million/3 Now plug it into the better equation to remedy for y y + 5x = 3 y = 3 - 5x y = 3 - 5(a million) y = 3 - 5(a million/3) y = -2 ; 4/3 so the recommendations are ( a million , -2 ) , ( a million/3 , 4/3)

Answer 6

x=y+4
substituting in the above equation i.e., x*x-2y=11
=>(y+4)(y+4)-2y=11
=>y*y+6y+5=0
=>y*y+y+5y+5=0
=>y*(y+1)+5*(y+1)=0
=>(y+1)*(y+5)=0
=>y+1=0 , y+5=0
=>y= -1, -5
but x=y+4
=>x=3 , -1

Answer 7

What ye has to do, my friend, is substitue in your formula for x into the top equation.

you have x=y+4, so x*x = (y+4)*(y+4)

expand the brackets and then youll have an equation in terms of y only. So then you can find the value of y.

Put this value for y into x=y+4 to find x.

Solved.

Answer 8

x*x-2y=11
x=y+4

(y+4)(y+4)-2y=11
y^2+8y+16-2y-11=0
y^2+6y+5=0
(y+5)(y+1)=0
y+5=0
y=-5,x=1

y+1=0
y=-1, x=3

Answer 9

well the answer is quite simple if you know what ti do first you tear up the sheet of paper then throw it away ....simple as that and you needed advise wow smart kid!

Answer 10

x = 3,-1
y = -1.-5