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Solve the given system of equations using the Substitution Method
( 0,-4), ( -sqrt(7), 3), and ( sqrt(7), 3)
We want to solve the given system of equations using the Substitution Method. x^2+y^2=16 & (I) y=x^2-4 & (II) The y-variable is isolated in Equation (II). This allows us to substitute its value x^2-4 for y in Equation (I).
(I): y= x^2-4
(I): (a+b)^2=a^2+2ab+b^2
(I): Calculate power and product
(I): LHS-16=RHS-16
(I): Subtract term
Notice that in Equation (I) we have a fourth order polynomial equation in terms of only the x-variable. Since there are no constant terms, it seems that trying to solve it by factoring will be a good idea. Let's do it!
Factor out x^2
Write as a power
a^2-b^2=(a+b)(a-b)
Use the Zero Product Property
(I): sqrt(LHS)=sqrt(RHS)
(II): LHS-sqrt(7)=RHS-sqrt(7)
(III): LHS+sqrt(7)=RHS+sqrt(7)
We found that we have three solution to Equation (I): x=0, x=-sqrt(7), and x=sqrt(7). We will substitute each of these values into Equation (II) to find the respective y-values of solutions to the original system. Let's start with x=0.
We found that one of the solutions to our system is point (0,-4). Now, let's substitute x=-sqrt(7) into Equation (II).
x= -sqrt(7)
(- a)^2=a^2
sqrt(a^2)=a
Subtract term
Therefore, the second solution is the point (-sqrt(7),3). Finally, let's substitute x = sqrt(7) into Equation (II).
x= sqrt(7)v
sqrt(a^2)=a
Subtract term
The last solution to our system is the point (sqrt(7),3).
(I), (II): x= 0, y= - 4
(I), (II): Calculate power
(I), (II): Add and subtract terms
Since both equations produced true statements, the solution (0,-4) is correct. Let's now check ( - sqrt(7), 3).
(I), (II): x= - sqrt(7), y= 3
(I), (II): Calculate power
(I), (II): Add and subtract terms
Since again both equations produce true statements, the solution ( - sqrt(7), 3) is also correct. Finally, let's check ( sqrt(7), 3).
(I), (II): x= sqrt(7), y= 3
(I), (II): Calculate power
(I), (II): Add and subtract terms
Once again both equations produce true statements. Therefore, ( sqrt(7), 3) is also correct.