System of equations ?:(

please solve the system

y^2 − 25x^2 =25
x= y^2 − 25

(x, y) = ( )
(x, y) = ( )
(x, y) = ( )
(x, y) = ( )

Dear unknown,

y^(2)-25x^(2)=25_x=y^(2)-25

Since the equation was solved for x, replace all occurrences of x in the other equations with the solution (y^(2)-25).
y^(2)-25(y^(2)-25)^(2)=25_x=y^(2)-25

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.
y^(2)+(-25(y^(4)-50y^(2)+625))=25_x=y^…

Multiply -25 by each term inside the parentheses.
y^(2)+((-25y^(4)+1250y^(2)-15625))=25_…

Remove the parentheses around the expression -25y^(4)+1250y^(2)-15625.
y^(2)+(-25y^(4)+1250y^(2)-15625)=25_x=…

Since y^(2) and 1250y^(2) are like terms, add 1250y^(2) to y^(2) to get 1251y^(2).
1251y^(2)-25y^(4)-15625=25_x=y^(2)-25

Reorder the polynomial 1251y^(2)-25y^(4)-15625 alphabetically from left to right, starting with the highest order term.
-25y^(4)+1251y^(2)-15625=25_x=y^(2)-25

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.
-25u^(2)+1251u-15650=0_x=y^(2)-25

Multiply each term in the equation by -1.
25u^(2)-1251u+15650=0_x=y^(2)-25

Find the factors such that the product of the factors is the trinomial 25u^(2)-1251u+15650. This can be done by trial and error and checked using the FOIL method of simplifying polynomials.
(u-25)(25u-626)=0_x=y^(2)-25

Set each of the factors of the left-hand side of the equation equal to 0.
u-25=0_25u-626=0_x=y^(2)-25

Since -25 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 25 to both sides.