Solve algebraically:
x^2 + y^2 = 18
x - 2y = -3.
Thanks.
To solve this system of equations algebraically, we can use the method of substitution or elimination. Let's use the substitution method in this case.
Step 1: Solve one of the equations for one variable.
Let's solve the second equation for x:
x - 2y = -3
x = 2y - 3
Step 2: Substitute the expression (2y - 3) for x in the first equation.
(x^2 + y^2) = 18
(2y - 3)^2 + y^2 = 18
Step 3: Simplify the equation.
(4y^2 - 12y + 9) + y^2 = 18
4y^2 - 12y + 9 + y^2 = 18
5y^2 - 12y + 9 = 18
Step 4: Rearrange the equation.
5y^2 - 12y + 9 - 18 = 0
5y^2 - 12y - 9 = 0
Step 5: Solve the quadratic equation.
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
In this case, the equation factors as:
(5y + 3)(y - 3) = 0
This gives two possible values for y:
y + 3/5 = 0
y = -3/5
or
y - 3 = 0
y = 3
Step 6: Substitute the found values of y back into the equation to solve for x.
Using y = -3/5:
x = 2(-3/5) - 3
x = -6/5 - 3
x = -6/5 - 3/1
x = -6/5 - 15/5
x = -21/5
Using y = 3:
x = 2(3) - 3
x = 6 - 3
x = 3
So, the solution to the system of equations is:
(x, y) = (-21/5, -3/5) and (3, 3).
To solve the given system of equations algebraically, we can use the method of substitution or elimination.
Let's use substitution to solve for one variable in terms of the other. From the second equation, we can isolate x:
x = -3 + 2y
Now, substitute this expression for x in the first equation:
(-3 + 2y)^2 + y^2 = 18
Expanding and simplifying this equation gives:
9 - 12y + 4y^2 + y^2 = 18
Combining like terms:
5y^2 - 12y - 9 = 0
To solve this quadratic equation, we can factor or use the quadratic formula. However, the equation does not have easy integer solutions. Therefore, we'll use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 5, b = -12, and c = -9. Substituting these values into the quadratic formula:
y = (-(-12) ± √((-12)^2 - 4(5)(-9))) / (2(5))
= (12 ± √(144 + 180)) / 10
= (12 ± √324) / 10
= (12 ± 18) / 10
So, we have two possible solutions for y:
y1 = (12 + 18) / 10 = 3
y2 = (12 - 18) / 10 = -0.6
Now, substitute these values back into the second equation to solve for x:
For y = 3:
x - 2(3) = -3
x - 6 = -3
x = -3 + 6
x = 3
For y = -0.6:
x - 2(-0.6) = -3
x + 1.2 = -3
x = -3 - 1.2
x = -4.2
Therefore, the solutions to the system of equations are:
(x, y) = (3, 3) and (-4.2, -0.6).
just substitute, since x=2y+3:
(2y+3)^2 + y^2 = 18
Solve that for y, and then x=2y+3