Given the following simultaneous equations, where a and b are non-zero, solve for x and y in terms of a,b,c?

Question

ax+by=c
a^2x^2-b^2y^2=-c^4

Answer 1

ax + by = c .......1

a²x² - b²y² = -c^4.......2

The 2nd equation can be written as

(ax+by)(ax-by) = -c^4

or c(ax-by) = -c^4 [as (ax+by) = c (eqn.1)]

or (ax-by) = -c³..........3

ADDING eqn 1 and eqn.3

2ax = c - c³ = c(1-c²)

or x = c(1-c²)/2a

AND SUBTRACTING eqn.3 from eqn.

2by = c + c³ = c(1+c²)

or y = c(1+c²)/2b

Answer 2

We rewrite the second equation thus:

(ax)^2 - (by)^2 = - c^4

(ax + by) (ax - by) = - c^4

c(ax - by) = - c^4

ax - by = - c^3

ax + by = c (given)

Add the two equations:

2ax = c - c^3

x = c (1 - c^2) / 2a

Similarly

y = -c (1 + c^2) / 2b

Answer 3

The clue is that the second equation is of the form



D squared - E squared which simplifies to (D-E)(D+E)



using the letters above the second equn. becomes



(ax-by)(ax+by)= c to power 4



First equn. tells us that (ax+by) = c



Thus

(ax-by) c = c power 4

ax-by = c power3

ax + by =c



Add these two



2ax = c + cpower 3

x = (c + c power3)/ 2a

Put this back in first equn. to get y = 1/b * [c- a * (c + c power3)/ 2a]

= [c -ac - ac power3] / 2b

=

Answer 4

Let's solve for "x" first. Solve the first equation for "y":

ax + by = c; subtract ax...

by = c - ax; divide by b...

y = (c - ax) / b



Now plug that value into our second equation:

a²x² - b²((c - ax) / b))² = c⁴; square both sides of the rational...

a²x² - (b²(c - ax)² / b²) = c⁴; reduce the rational...

a²x² - (c - ax)² = c⁴; square...

a²x² - (c² - 2acx + a²x²) = c⁴; eliminate parentheses...

a²x² - c² + 2acx - a²x² = c⁴; combine like terms...

-c² + 2acx = c⁴; add c²...

2acx = c⁴ + c²; divide by 2ac...

x = (c⁴ + c²) / 2ac; reduce...

x = (c³ + c) / 2a



Use a similar process to solve for "y". First, solve the first equation for "x":

ax + by = c; subtract by...

ax = c - by; divide by a...

x = (c - by) / a



Substitute in the second equation:

a²((c - by) / a)² - b²y² = c⁴; square the rational...

(a²(c - by)² / a²) - b²y² = c⁴; reduce...

(c - by)² - b²y² = c⁴; square...

c² - 2bcy + b²y² - b²y² = c⁴; combine like terms...

c² - 2bcy = c⁴; subtract c²...

-2bcy = c⁴ - c²; divide by -2bc...

y = (c⁴ - c²) / -2bc; reduce...

y = (c³ - c) / -2b

Answer 5

ax+by=c



a^2x^2-b^2y^2=-c^4 rewrite like

(ax+by)(ax-by)=-c^4 substitute first eq.

c(ax-by)=-c^4

obtain

ax-by = -c^3

ax+by = c

add them

2ax=c(1-c^2)

x=c(1-c^2)/(2a)

subtract them

2by =c(1+c^2)

y=c(1+c^2)/(2b)

Answer 6

applying A^2 - B^2 formula you get



(ax+by)(ax-by) = -c^4



so

c*(ax-by) = -c^4



ax-by = -c^3



now add 1 + this eq and subtract 1+ this eq



2ax = c-c^3

2by = c+c^3



x = (c-c^3)/2a y = (c+c^3)/2b

Answer 7

ax+by potential of=c ie, x=(c-by potential of)/a substituting in a^2x^2-b^2y^2=-c^4 a^2 (c-by potential of/a)^2 - b^2.y^2 = -c^4 ie, a^2 (c^2+b^2y^2-2bcy)/a^2 - b^2y^2=-c^4 ie, c^2 + b^2y^2 -2bcy =-c^4 ie, b^2.y^2 - 2bcy +c^2 + c^4=0 ie, y=[2bc +- sqrt(4b^2.c^2-4(b^2)(c^4))]/2b^2 ie, y=[2bc +- 2bc.sqrt(a million-c^2)]/2b^2 ie, y=2bc(a million+-sqrt(a million-c^2))/2b^2 ie, y=(c/b)(a million+-sqrt(a million-c^2)) substituting this value of y in ax+by potential of=c: ax+b(c/b)(a million+-sqrt(a million-c^2))=c ie, ax+c.(a million+-sqrt(a million-c^2))=c ie, ax=c-c.(a million+-sqrt(a million-c^2)) ie, x=(c/a)(-+sqrt(a million-c^2) ie, x=-+(c/a).sqrt(a million-c^2) so, x=-+(c/a).sqrt(a million-c^2) and y=(c/b)(a million+-sqrt(a million-c^2))