Help needed with systems of equations?

Question

(a - b)x + (a + b)y = a^2 - 2ab - b^2
(a + b)(x + y) = a^2 - b^2

There is a second problem:

px + qy = p - q
qx - py = p + q

Both are simultaneous equations in x and y.

You just need to find x and y in terms of the constants. Yes, the alphabets apart from x and y are constants.

a,b,p and q are constants. You need to express x and y in terms of those constants, like x = 2p/q and the sort

The (a+b)x and the px + qy anre separate questions!. The problem in Additional details is the second question. The one in Original details is the first question. Don't solve them together!

Answer 1

(a – b) x + (a + b) y = a2 – 2ab – b2(equation 1)
(a + b)(x + y) = a^2 – b^2 (equation 2)

Step 1open the second bracket in the second equation and keep the first equation as it is

(a – b) x + (a + b) y = a^2 – 2ab – b^2 (equation 1)
(a + b) x + (a + b) y = a^2 – b^2(equation 2)

Step 2 Subtract the second equation from the first

(a – b) x – (a + b) x = –2ab

Step 3 Simplify

- 2bx = - 2ab
x = a
Step 4 Substitute this value of x in equation 2 and simplify

(a + b)(a + y) = a^2 – b^2
(a + b) (a + y) = (a + b) (a – b) Here we have factorized a^2 – b^2

Divide both sides by (a + b). We assume that a + b is not zero

(a + y) = (a – b)
y = –b
Now try the second set of equations on your own.
I will give you a hint; multiply the first equation by p and the second equation by q

Answer 2

E1: 2x + y = -2 E2: -x + y = 11 the assumption is to function or subtract sides so the coefficient of one of the variables will become 0. To get rid of x it is mandatory to multiply E2 on the two sides by potential of two so the coefficients of x are opposites. E3 = E2 + 2*E1 2x + y + 2(-x + y) = -2 + 2(11) 2x + y + -2x + 2y = -2 + 22 0x + 3y = 20 y = 20/3 = 6 2/3 = 6.66666... So 2x + 20/3 = -2 2x = -2 – 20/3 = -26/3 x = -26/3*a million/2 = -13/3 = -4 a million/3 = -4.333333... examine those outcomes in the unique equations, I did. ProfRay

Answer 3

Well, the basic rule involving a series of equations requires the number of equations to equal the number of unknowns. So for either of these systems, you need to find two more equations.

Answer 4

2nd problem: px+qy=p-q,qx-py=p+q

from qx-py=p+Q
-py= -qx+p+q
-py= -(qx-p-y)
py=qx-(p+y)
y=[qx-(p+q)]/p
putting this value of y in 1st eq.
px+[q(qx-(p+q)]/p=p-q
px+[q^2x-pq-q^2]/p=p-q
p^2x+q^2x-pq-q^2=p^2-pq
p^2x+q^2x-q^2=p^2=q^2
x(p^2+q^2)=p^2+q^2
x= (p^2+q^2)/(p^2+q^2)
x= 1
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putting this value of x in any of the eq. you will get value of y
we put x=1 in eq.1
p*1 +qy=p-q
p+qy=p-q
qy=p-q-p
qy= -q
y= -1.
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hence x=1 and y= -1.
the first problem u hav given is according 2 me wrong.check once.
well i am a student of 9th class and these things r in my course.
bye.

Answer 5

let (a+b)=p,(a-b)=q
substitute
1. qx +py = (a-b)(a-b)= qq
2. px +py = (a+b)(a-b)=pq
from eqn 2, x= (pq-py)/p
substitute in eqn 1 solves for y= p-q
substitute y back in eqn 1 solves for x=-p+2q
substitute for p and q, then y=2b and x=a-3b