Useing substitution/elimination method what is the solution set to x^2+y^2=25
4x-3y=0
a)(3,4) and (-3,-4)
b)(3,-4) and (-3,4)
c)(-4,3) and (4,-3)
d)(4,3) and (-4,-3)
PLEASE HELP!
There are two ways to do this.
(1) Try each of the multiple choices in the equations and see which one works, or
(2) Substitute 4x/3 for y in the first equation, making it an equation for x only. Solve that. With each of the x solutions you get, compute y.
Note that, since y = (4/3) x, x and y must have the same sign. This eliminates choices (b) and (c) right away. A bit more thought should help you eliminate (d) as an option.
To solve this system of equations using the substitution method, we can start by solving one equation for one variable and then substitute that expression into the other equation.
Given:
Equation 1: x^2 + y^2 = 25
Equation 2: 4x - 3y = 0
Let's solve Equation 2 for x:
4x - 3y = 0
4x = 3y
x = (3/4)y
Now, substitute x = (3/4)y into Equation 1:
(x^2) + y^2 = 25
((3/4)y)^2 + y^2 = 25
(9/16)y^2 + y^2 = 25
(9/16)y^2 + (16/16)y^2 = 25
(25/16)y^2 = 25
y^2 = (25 * 16) / 25
y^2 = 16
Taking the square root of both sides, we get:
y = ±4
Now, substitute the values of y back into x = (3/4)y to find the corresponding values of x:
For y = 4:
x = (3/4)(4)
x = 3
For y = -4:
x = (3/4)(-4)
x = -3
So the solutions to the system of equations are:
a) (3, 4) and (-3, -4)
Therefore, the solution set is (a) (3, 4) and (-3, -4).