Isolate one variable in Equation (II). Then, substitute the variable's equivalent value into Equation (I).
(- 2,2), ( 2,- 2 )
Practice makes perfect
We will solve the given system of equations using the Substitution Method.
x^2+y^2=8 & (I) x+y=0 & (II)
Note that neither of the variables is isolated in either equation, so we need to start by isolating x in Equation (II).
x^2+y^2=8 x+y=0 ⇔ x^2+y^2=8 x=- y
The x-variable is isolated in Equation (II). This allows us to substitute its value - y for x in Equation (I).
Now, consider Equation (II).
x=- y
We can substitute y=2 and y=- 2 into the above equation to find the values for x. Let's start with y= 2.
x=- y ⇒ x=- 2
We found that x=- 2 when y=2. One solution of the system is (- 2,2). To find the other solution, we will substitute - 2 for y in Equation (II) again.
We found that x=2 when y=- 2. Therefore, our second solution is ( 2,- 2 ).
Checking Our Answer
Checking the answer
We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (- 2,2). We will substitute - 2 and 2 for x and y, respectively, in Equation (I) and Equation (II).
