simultaneous equations :(?

Question

I have just sat one of my maths papers in school and I am so irritated because the simultaneous equations had a squared x value in it!! And I was completely lost! It was:

y=5x-1
y=2x^2+1

Please help me it was horrible!!

By the way it was non calculator.
It was AQA exam board...strange how it was the same question :S

Answer 1

y=5x-1 y=2x^2+1
then 5x-1=2x^2+1
2x^2+1-5x+1=0
2x^2-5x+2=0
(2x-1)(x-2)=0 ................x = 1/2 .....y = 5(1/2) - 1 = 3/2
or x =2 .........y = 5(2) - 1 = 9

Answer 2

y=5x - 1
y=2x² + 1

5x - 1 = 2x² + 1

Combine terms

2x² - 5x + 2 = 0

factorise:

(x - 2)(x - ½) = 0

x = 2, x = ½

Substitute the values of x to find y:
if x = 2, y = 5(2) - 1 = 9
if x = ½, y = 5(½) - 1 = 4½

Answer 3

Looks like a trick question but its really easy.

as you can see, both equation are equal to y. therefore you can equate them: 5x-1=2x^2+1

rearranging this: 2x^2-5x+2=0
you solve this quadratic equation to give x=1/2 and x=2

when x=1/2, y=5(1/2)-1=1.5
when x=2, y=5(2)-1= 9

Answer 4

there was no squared x in it. all you had to do was to make the two equations simular. You do this by making either x or y the same. so by this i would suggest making it 10x squared to x. not forgetting to do the same to the rest of the equation. One of the 10x needs to be a negative because it cancels the other out leaving you with y=somthing. What the hell is the receprocal of 2. what does receprecol even mean? i got to admit it was alright until the last 2 pages. Dont worry i am sure you did fine. Your teacher would not of put you on that exam if they thought you couldn't do it. Hope that makes you feel a bit better.

Answer 5

You just equate these:

As you know two values for y, you can use this to get:

5x-1=2x^2+1

From that rearrange to:
2x^2-5x+2=0

Which is a quadratic equation, factorising gives:
(2x-1)(x-2)=0

so x=2 or x=0.5

Now, you can plug these x values into either equation:

y=5(2)-1=9
y=5(0.5)-1=1.5

Answer 6

since y = 5x-1 and y = 2x^2+1, you can substitute the value of y in one equation for y in the other:
5x-1 = 2x^2+1
move everything to one side of the equal sign:

2x^2-5x+2 = 0

using the quadratic equation: x = (-b +/- (b^2-4ac)^1/2)/2a;
a=, b=-5, c=2, you'll find that

x=2, .5

Answer 7

y=5x-1.....(equation 1)
y=2x^2+1.....(equation 2)
subtract equation 1 from 2
y-y=2x^2+1-5x+1
0=2x^2-5x+2
2x^2-4x-x+2=0
2x(x-2)-1(x-2)=0
x-2=0 or 2x-1=0
x=2 or 1/2
substitute value of x in any of the equation
y=5(2)-1
y=10-1=9
y=5(1/2)-1=3/2
y=9 or3/2

Answer 8

y=5x - 1
y=2x² + 1

5x - 1=2x² + 1

Combine terms

2x² - 5x +2 = 0

x² - (2½)x +1 = 0

factor

(x - 2)(x - ½) = 0

x = 2, x = ½
.

Answer 9

I tried to take the first derivative dy/dx, but it doesn't work.
y'=5
y'=4x
x=5/4
y=5(5/4)-1=5.25
y=2(5/4)^2+1=4.125

ok try this
5x-1=2x^2+1 ;(y=y)
2x^2-5x+2=0
now apply quadratic formula
x1=(5+sqrt(25-16))/4
x1=5/4 + sqrt(9/16)
x2=5/4 - sqrt(9/16)

Answer 10

2x² + 1 = 5x - 1
2x² - 5x + 2 = 0
(2x - 1).(x - 2) = 0
x = 1 / 2 , x = 2