solve the following system of equations.
2 x (square root of x) +5y=16
6 x (square root of x) - 3y = 12
To solve this system of equations, we can use the method of substitution.
First, we will solve one equation for one variable and substitute it into the other equation. Let's solve the first equation for x:
2√x + 5y = 16
2√x = 16 - 5y
√x = (16 - 5y) / 2
x = [(16 - 5y) / 2]^2
x = (16 - 5y)^2 / 4
Now, substitute this expression for x into the second equation:
6√x - 3y = 12
6[(16 - 5y)^2 / 4] - 3y = 12
[(16 - 5y)^2 / 2] - 3y = 12
(16 - 5y)^2 - 6y = 24
(16 - 5y)^2 = 24 + 6y
Now, we have a quadratic equation in terms of y. Let's simplify it further:
(16 - 5y)^2 = 24 + 6y
256 - 160y + 25y^2 = 24 + 6y
25y^2 - 166y + 232 = 0
This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula.
Let's solve it by factoring:
25y^2 - 166y + 232 = 0
(5y - 8)(5y - 29) = 0
Setting each factor equal to zero, we get:
5y - 8 = 0 or 5y - 29 = 0
Solving each equation for y, we find:
5y = 8 or 5y = 29
y = 8/5 or y = 29/5
Now, substitute these values of y back into the first equation to find x.
When y = 8/5:
2√x + 5(8/5) = 16
2√x + 8 = 16
2√x = 8
√x = 4
x = 4^2
x = 16
When y = 29/5:
2√x + 5(29/5) = 16
2√x + 29 = 16
2√x = 16 - 29
2√x = -13 (This step indicates that there are no real solutions for x when y = 29/5)
Therefore, the solution to the system of equations is:
x = 16 and y = 8/5
To solve the given system of equations:
Step 1: Let's start by simplifying the equations. Since both equations contain the square root of x, let's substitute a variable to represent it. Let's say u = √x. Now, we can rewrite the system of equations as:
2u + 5y = 16 -- Equation 1
6u - 3y = 12 -- Equation 2
Step 2: We can solve Equation 2 for u by isolating it. Divide both sides of Equation 2 by 6:
(6u - 3y)/6 = 12/6
u - 0.5y = 2 -- Equation 3
Step 3: Now we have two equations. We can eliminate u by multiplying Equation 3 by 2:
2(u - 0.5y) = 2(2)
2u - y = 4 -- Equation 4
Step 4: Now we have two equations in terms of y. We can solve for y by subtracting Equation 1 from Equation 4:
(2u - y) - (2u + 5y) = 4 - 16
2u - y - 2u - 5y = -12
Simplifying:
-6y = -12
Divide both sides by -6:
y = 2
Step 5: Now that we have found the value of y, substitute it back into Equation 1 to find u:
2u + 5(2) = 16
2u + 10 = 16
2u = 16 - 10
2u = 6
Divide both sides by 2:
u = 3
Step 6: Recall that u = √x. Substitute the value of u to determine x:
3 = √x
Square both sides:
9 = x
Step 7: Therefore, the solution to the system of equations is x = 9 and y = 2.
To solve this system of equations, we will use a technique called substitution. Firstly, let's solve the first equation for x in terms of y.
2√x + 5y = 16
Subtract 5y from both sides:
2√x = 16 - 5y
Divide both sides by 2:
√x = (16 - 5y) / 2
Square both sides to eliminate the square root:
x = ((16 - 5y) / 2)^2
Now we have an expression for x in terms of y that we can substitute into the second equation. Let's substitute x in the second equation:
6√x - 3y = 12
Replace x with its expression:
6√((16 - 5y) / 2)^2 - 3y = 12
Simplify the expression inside the square root:
6((16 - 5y) / 2) - 3y = 12
Simplify further:
3(16 - 5y) - 3y = 12
Distribute the 3:
48 - 15y - 3y = 12
Combine like terms:
48 - 18y = 12
Subtract 48 from both sides:
-18y = -36
Divide both sides by -18:
y = 2
Now that we have found the value of y, we can substitute it back into the first equation to find x. Using the expression we found for x earlier:
x = ((16 - 5(2)) / 2)^2
x = ((16 - 10) / 2)^2
x = (6 / 2)^2
x = 3^2
x = 9
Therefore, the solution to the system of equations is x = 9 and y = 2.