How to solve for x and y using the simultaneous equation method

3y + 4x = 25, x^2 + y^2 = 25.

Please show me the full steps to solve this using the any simultaneous equation method (substitution, elimination, addition, etc). Thanks.

solve the first equation for y in term of x:
3y + 4x = 25
y = (25-4x)/3
y² = (25-4x)²/9

Substitute for y² in the second equation:
x² + (25-4x)²/9 = 25
9x² + (25-4x)² = 225

Now expand this and you'll have a quadratic equation in x which you can solve for x (2 values probably). Substitute each of those values of x into the first equation and solve for the corresponding value of y.

9x² + 625 - 200x + 16x² = 225
25x² - 200x + 400 = 0
x² - 8x + 16 = 0
(x-4)² = 0
x = 4 (multiplicity 2)
3y + 4(4) = 25
3y = 25-16 = 9
y = 3

Solution (4,3)
The two equations describe a circle of radius 5 centered on the origin, and a line tangent to the circle at the point (4, 3). There is only one distinct solution because the line is tangent at a point instead of intersecting the circle at two points.

And there you have it.

Solve for x in terms of y, and then plug that value into the equation x^2 + y^2 = 25.
y = (25-4x)/3
x^2 + ((625 - 200x + 16x^2)/9) = 25
Multiply the whole equation by 9:
9x^2 + 625 - 200x + 16x^2 = 225
25x^2 - 200x - 400 = 0
Divide the equation by 25:
x^2 - 8x - 16 = 0
Factor the equation:
(x-4)(x-4) = 0
x = 4

Plug into 3y + 4x = 25 to solve for y.
3y + 4(4) = 25
3y = 25 - 16 = 9
y = 3

So x = 4 and y = 3.

first equation is 3x+4y=25
now x=(25-4y)/3
Substitute the value of x in the second equation x^2+y^2=25
we will get the value of Y, might be 2 coz this is quadratic equation. then substitute the value of y we will ge the value of x.