1. equation

hey i have yet another maths question... i tried looking for the solution but the site for the exam papers is down. the question is solve:

for x and y
x + y = 7

x^2+y^2=29

the answer is (2,5) and (5,2) but i got (10,0) and (0,10)

i have ALOT of difficulty with these sums so i would be really grateful for the solution to do it , thanks

then it asks in part (2) " which one of the values of y in the above satisfies the inequality 6-2y<0

2. Originally Posted by cliste09
hey i have yet another maths question... i tried looking for the solution but the site for the exam papers is down. the question is solve:

for x and y
x + y = 7

x^2+y^2=29

the answer is (2,5) and (5,2) but i got (10,0) and (0,10)

i have ALOT of difficulty with these sums so i would be really grateful for the solution to do it , thanks

then it asks in part (2) " which one of the values of y in the above satisfies the inequality 6-2y<0
Hi cliste09,

(1) $x+y=7$
(2) $x^2+y^2=29$

Solve the (1) equation for x
(1) $x=7-y$

Substitute into (2)

$(7-y)^2+y^2=29$
$49-14y+y^2+y^2-29=0$
$2y^2-14y+20=0$
$y^2-7y+10=0$
$(y-5)(y-2)=0$
$y=5 \ \ or \ \ y=2$

Substitute each value of y back into (1) to retrieve your x-values. Your points of intersection are (2,5) and (5, 2).

As far as part 2 goes, use your y-values of 5 and 2 and see which one makes the inequality true.

3. thank you soooo much!!!!! your solution is very easy to understand , phew these q's are one of my worst areas in maths ... dont even get me started on trigonometry lol