Solve the system x+y=9, x^2 - y^2 = 36 for x and y

  • Thread starter Thread starter aricho
  • Start date Start date
  • Tags Tags
    System
Click For Summary
SUMMARY

The discussion revolves around solving the system of equations: x + y = 8 and x^2 - y^2 = 36. Participants clarify that the correct approach involves substituting y with (8 - x) in the second equation, leading to the equation 16x - 64 = 36. This simplifies to x = 6.25 and y = 1.75. The final values satisfy both original equations, confirming the solution is correct.

PREREQUISITES
  • Understanding of algebraic equations and systems of equations
  • Familiarity with substitution methods in solving equations
  • Knowledge of factoring techniques, particularly with differences of squares
  • Basic skills in manipulating and simplifying algebraic expressions
NEXT STEPS
  • Study methods for solving systems of linear equations, including substitution and elimination
  • Learn about the difference of squares and its applications in algebra
  • Explore quadratic equations and their solutions using the quadratic formula
  • Practice solving word problems that involve systems of equations
USEFUL FOR

Students preparing for mathematics exams, educators teaching algebra, and anyone looking to improve their problem-solving skills in algebraic contexts.

aricho
Messages
71
Reaction score
0
Hi, i just sat my maths exam, and i had a very strange question... i don't know how to do it...

x+y=8
x^2-y^2=36

find the values for x and y

Thanks
 
Physics news on Phys.org
y = 8-x

x^2 - (8-x)^2 = 36

x^2 - (64-16x+x^2) = 36

16x - 64 = 36
 
is that it?

i did y^2-x^2=36
(x-y)(x+y)=36
(x-y)(8)=36
x-y=36/8

is any of that correct?
 
According to Whozum, 16x- 64= 36 so x= 100/16= 25/4= 6.25 and y= 8- 6.25= 1.75.

x-y= 6.25- 1.75= 4.5 and 36/8= 9/2= 4.5.

Your statement is correct but is not a solution to the "problem" which, I suppose, was to solve the two equations.

In fact, the only thing strange I see about your "question" is that there was no "question"! Are you sure you didn't leave something out- like "solve this pair of equations" or "what are x and y"?
 
aricho said:
is that it?

i did y^2-x^2=36
(x-y)(x+y)=36
(x-y)(8)=36
x-y=36/8

is any of that correct?
Yup. This is correct. Since you have 2 unknowns, you need 2 equations. And you have already had 2 equations. You can then solve:
\left\{ \begin{array}{l}x + y = 8 \\ x - y = \frac{9}{2} \end{array} \right.
for x, and y.
Viet Dao,
 
HallsofIvy said:
Your statement is correct but is not a solution to the "problem" which, I suppose, was to solve the two equations.

Me or him?
 

Similar threads

Replies
8
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 40 ·
2
Replies
40
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
837
  • · Replies 25 ·
Replies
25
Views
2K
Replies
11
Views
1K