Solve alebratically the simultaneous equation
x^2 + y^2 = 25
y - 2x = 5
please explain every step
Here's how to do it using substitution. There might be a quicker way though....
The problem tells us...
y - 2x = 5 Add 2x to both sides of this equation.
y = 5 + 2x
The problem tells us...
x2 + y2 = 25 And since y = 5 + 2x , we can replace y with 5 + 2x .
x2 + (5 + 2x)2 = 25
x2 + (5 + 2x)(5 + 2x) = 25 Multiply out the parenthesees.
x2 + (5)(5) + (5)(2x) + (2x)(5) + (2x)(2x) = 25
x2 + 25 + 10x + 10x + 4x2 = 25 Combine like terms.
5x2 + 25 + 20x = 25 Subtract 25 from both sides of the equation.
5x2 + 20x = 0 Factor out an x from both terms.
x(5x + 20) = 0 Set each factor equal to 0 and solve for x .
x = 0 or 5x + 20 = 0 Subtract 20 from both sides.
5x = -20 Divide both sides by 5 .
x = -4
Now plug these values for x into the second equation given. (The first one will give you two answers for y , but only one answer for y works in the second equation.)
y - 2x = 5 Plug in 0 for x . y - 2(0) = 5 y - 0 = 5 y = 5 | y - 2x = 5 Plug in -4 for x . y - 2(-4) = 5 y - -8 = 5 y + 8 = 5 Subtract 8 from both sides. | |
y = -3 |
So the two solutions are:
x = 0, y = 5 and x = -4, y = -3
Here's how to do it using substitution. There might be a quicker way though....
The problem tells us...
y - 2x = 5 Add 2x to both sides of this equation.
y = 5 + 2x
The problem tells us...
x2 + y2 = 25 And since y = 5 + 2x , we can replace y with 5 + 2x .
x2 + (5 + 2x)2 = 25
x2 + (5 + 2x)(5 + 2x) = 25 Multiply out the parenthesees.
x2 + (5)(5) + (5)(2x) + (2x)(5) + (2x)(2x) = 25
x2 + 25 + 10x + 10x + 4x2 = 25 Combine like terms.
5x2 + 25 + 20x = 25 Subtract 25 from both sides of the equation.
5x2 + 20x = 0 Factor out an x from both terms.
x(5x + 20) = 0 Set each factor equal to 0 and solve for x .
x = 0 or 5x + 20 = 0 Subtract 20 from both sides.
5x = -20 Divide both sides by 5 .
x = -4
Now plug these values for x into the second equation given. (The first one will give you two answers for y , but only one answer for y works in the second equation.)
y - 2x = 5 Plug in 0 for x . y - 2(0) = 5 y - 0 = 5 y = 5 | y - 2x = 5 Plug in -4 for x . y - 2(-4) = 5 y - -8 = 5 y + 8 = 5 Subtract 8 from both sides. | |
y = -3 |
So the two solutions are:
x = 0, y = 5 and x = -4, y = -3