How in steps would I work out this;

x^2 + y^2 = 5

3x + y = 7

Thanks loads for any help :)

Posted Sun 17th February, 2013 @ 16:57 by

:) PurpleJaguar (: - Team GR
First you rearrange the second equation to make this:

y=7-3x

then you put replace every y in the first eqaution with (7-3x)

x^2 + (7 - 3x)(7 - 3x) = 5

then you expand the brackets:

x^2 + 49 - 42x +9x^2 = 5

then you take the 5 to the otherside and simplify

10x ^2 - 42x + 44 =0

then you factorise

so times number in front of x^2 which is 10 by the number at the end which is 44 (this gives you 440).

Then find two numbers that times together to 440 and add to make - 42 (which are -22 and -20)

Then replace the -42x part of the equation with -22x and -20x...

10 x^2 -20x - 22x + 44 = 0

then you factorise (btw the expression in both brakets have to be the same)

10x (x - 2) -22(x - 2) = 0

(10 x - 22)(x - 2)= 0

then split them up

10 x - 22 = 0

x = 22/10

x = 2.2

or

x - 2 = 0

x = 2

then you put these to values of x in the very first equation

3x + y = 7

3(2.2) + y = 7

y = 7 - 6.6

y = 0.4

and for the other value of x

3(2) + y = 7

y = 1

so the answers are:

x= 2 and y = 1

or

x= 2.2 and y= 0.4

(O.O that was looooong sry)

(btw if its a calculator exam just put it in the quadratic equation formula from the point 10x ^2 - 42x + 44 =0, then put the two x values you get into the first or secound eqation that they gave you in the question)

Answered Mon 18th February, 2013 @ 14:27 by

gloria
Wow, thanks a lot, this is really really helpful and thanks for taking (probably)loads of your time to do it

Thanks again

p.s yeah its a calc paper

Would this be A* stuff for GCSE?

Answered Mon 18th February, 2013 @ 17:22 by

:) PurpleJaguar (: - Team GR
np, and i think it would be an A/A* question(for GCSE)

Answered Tue 19th February, 2013 @ 20:30 by

gloria
Answered Wed 20th February, 2013 @ 17:28 by

:) PurpleJaguar (: - Team GR