Solve the system by substitution or elimination.
x^2 + 2y^2 = 8
x^2 - y^2 = 1
subtract the equations to get
3y^2 = 7
y = ±√(7/3)
so,
x^2 - 7/3 = 1
x^2 = 10/3
x = ±√(10/3)
I'm sure you'd have had no trouble if it were
x+2y=8
x-y=1
Thank you! I tried to solve it like this:
x^2 + 2y^2 = 8
( x^2 - y^2 = 1 )2
x^2 + 2y^2 = 8
2x^2 - 2y^2 = 2
3x^2 = 10
x^2 = 10/3
x = ±√(10/3)
Any idea as to why my textbook would say that the solution is ±√(30/3)?
I figured it out. No fraction can appear under a radical.
Check the bracket placement.
±√(10/3)
= ±√(10)/√(3)
= ±√(30)/3
= ±(⅓)√(30)
Yeah, lots of folks don't like to see radicals in the denominator. Pretty much a non-issue with me, but since it's a common practice, be sure to watch for it.
To solve the system of equations by substitution or elimination, we will first rearrange one equation to solve for one variable and then substitute this expression into the other equation. Let's solve the second equation for x^2.
Given:
Equation 1: x^2 + 2y^2 = 8
Equation 2: x^2 - y^2 = 1
Step 1: Solve the second equation for x^2.
Rearrange Equation 2: x^2 = y^2 + 1
Now, we have an expression for x^2. We can substitute this expression into the first equation.
Step 2: Substitute the expression for x^2 into the first equation.
Substitute x^2 = y^2 + 1 into Equation 1:
(y^2 + 1) + 2y^2 = 8
Combine like terms:
3y^2 + 1 = 8
Subtract 1 from both sides:
3y^2 = 7
Divide by 3:
y^2 = 7/3
Now that we have the value of y^2, we can substitute it back into Equation 2 or Equation 1 to find the values of x.
Step 3: Solve for y.
Take the square root of both sides:
y = ±√(7/3)
Step 4: Substitute y into either equation to solve for x.
Let's use Equation 2 to find x.
x^2 - y^2 = 1
Substitute the value of y:
x^2 - (7/3) = 1
Multiply through by 3 to clear the fraction:
3x^2 - 7 = 3
Add 7 to both sides:
3x^2 = 10
Divide by 3:
x^2 = 10/3
Step 5: Solve for x.
Take the square root of both sides:
x = ±√(10/3)
Therefore, the solutions to the system of equations are:
x = ±√(10/3) and y = ±√(7/3)