Solve the system of equations and check your solution.
x^2+y^2=25
x+2y=10
I tried but I got only the 2 x values 8 plus or minus the square root of 164 all over 2.
I know the answer is wrong but I can't find my mistake.
Can you show me step-by-step how to solve this system of equation?
Just a fun fact: the first equation is a circle with radius of 5 (square root of 25).
I'll use substitution to answer this.
Let's start with defining x: x=10-2*y
Insert into the first equation: (10-2*y)^2+y^2=25
Expand the equation: 100-20y-20y+4y^2+y^2=25
Simplify: 5y^2-40y+100=25
Simplify: 5y^2-40y+75=0
Divide both sides by 5:y^2-8y+15=0
Solve for 0 using the algebraic formula:
(8+(sqrt(8^2-4*1*15)))/2*1
=(8+sqrt(4))/2; sqrt(4)=2
and (8-sqrt(4))/2
So the answers are 10/2 and 6/2 or y=3 and y=5
Plug these back into the second equation:
x+2y=10
x+2*3=10 and x+2*5=10
x+6=10 and x+10=10
x=4 and x=0
You probably just made a multiplication error inside the square root.
You can check the answer by graphing the two and seeing where they intersect. I use Desmos for graphing.
x + 2 y = 10 Subtract x to both sides
x + 2 y - x = 10 - x
2 y = 10 - x Divide both sides by 2
2 y / 2 = 10 / 2 - x / 2
y = 5 - x / 2
x ^ 2 + y ^ 2 = 25
x ^ 2 + ( 5 - x / 2 ) ^ 2 = 25
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Remark:
( a - b ) ^ 2 = a ^ 2 - 2 * a * b + b ^ 2
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x ^ 2 + [ 5 ^ 2 - 2 * 5 * x / 2 + ( x / 2 ) ^ 2 ] = 25
x ^ 2 + 25 - 5 x + x ^ 2 / 4 = 25
x ^ 2 + x ^ 2 / 4 + 25 - 5 x = 25
4 x ^ 2 / 4 + x ^ 2 / 4 + 25 - 5 x = 25
5 x ^ 2 / 4 + x ^ 2 + 25 - 5 x = 25 Subtract 25 to both sides
5 x ^ 2 / 4 + 25 - 5 x - 25 = 25 - 25
5 x ^ 2 / 4 - 5 x = 0 Divide both sides by 5
x ^ 2 / 4 - x = 0
x ( x / 4 - 1 ) = 0
Split into two equations.
x = 0
and
x / 4 - 1 = 0 Add 1 to both sides
x / 4 - 1 + 1 = 0 + 1
x / 4 = 1 Multiply both sides by
x = 4
Replace x = 0 and x = 4 into equation x + 2 y = 10
x = 0
0 + 2 y = 10
2 y = 10 Divide both sides by 2
y = 5
x = 4
4 + 2 y = 10 Subtract 4 to both sides
4 + 2 y - 4 = 10 - 4
2 y = 6 Divide both sides by 2
y = 3
The solutions are :
x = 0 , y = 5
and
x = 4 , y = 3
You can write solutions like :
( 0 , 5 ) and ( 4 , 3 )
Chek solutions :
x = 0 , y = 5
x ^ 2 + y ^ 2 = 25
0 ^ 2 + 5 ^ 2 = 25
0 + 25 = 25
25 = 25 Correct
x + 2 y = 10
0 + 2 * 5 = 10
0 + 10 = 10
10 = 10 Correct
x = 4 , y = 3
x ^ 2 + y ^ 2 = 25
4 ^ 2 + 3 ^ 2 = 25
16 + 9 = 25
25 = 25 Correct
x + 2 y = 10
4 + 2 * 3 = 10
4 + 6 = 10
10 = 10 Correct
To solve the system of equations, let's use the method of substitution:
Step 1: Solve one equation for one variable in terms of the other variable.
Let's solve the second equation for x:
x + 2y = 10
x = 10 - 2y
Step 2: Substitute the expression for x from Step 1 into the other equation.
Substituting x = 10 - 2y into the first equation:
(10 - 2y)^2 + y^2 = 25
Simplifying:
100 - 40y + 4y^2 + y^2 = 25
Combine like terms:
5y^2 - 40y + 75 = 0
Step 3: Solve the quadratic equation obtained in Step 2.
To solve this quadratic equation, you can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
The quadratic formula states that the solutions to the equation ax^2 + bx + c = 0 are given by:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our equation 5y^2 - 40y + 75 = 0, the coefficients are a = 5, b = -40, and c = 75.
Substituting these values into the quadratic formula:
y = (-(-40) ± sqrt((-40)^2 - 4(5)(75))) / (2(5))
Simplifying inside the square root:
y = (40 ± sqrt(1600 - 1500)) / 10
y = (40 ± sqrt(100)) / 10
y = (40 ± 10) / 10
This gives us two possible values for y:
1. y = (40 + 10) / 10 = 50 / 10 = 5
2. y = (40 - 10) / 10 = 30 / 10 = 3
Step 4: Find the corresponding values of x.
To find the values of x, substitute these values of y back into one of the original equations. Let's use the second equation:
x + 2y = 10
For y = 5:
x + 2(5) = 10
x + 10 = 10
x = 0
So one solution is x = 0, y = 5.
For y = 3:
x + 2(3) = 10
x + 6 = 10
x = 4
So the other solution is x = 4, y = 3.
Step 5: Check the solution.
To check if these solutions are correct, substitute the values of x and y into both equations and see if they satisfy the given equations.
For x = 0, y = 5:
x^2 + y^2 = 0^2 + 5^2 = 0 + 25 = 25 (satisfied)
x + 2y = 0 + 2(5) = 10 (satisfied)
For x = 4, y = 3:
x^2 + y^2 = 4^2 + 3^2 = 16 + 9 = 25 (satisfied)
x + 2y = 4 + 2(3) = 4 + 6 = 10 (satisfied)
Therefore, the solutions to the system of equations are x = 0, y = 5 and x = 4, y = 3.