Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 7.1
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Exercise 11 Page 401

Practice makes perfect
a We want to solve the given system of equations, if we can.

y=-2x-1 & (I) y= 12x-16 & (II) At least one of the variables has a coefficient of 1, so let's use the substitution method to solve it. When solving a system of equations using substitution, we follow three steps.

  1. Isolate a variable in one of the equations.
  2. Substitute the expression for that variable into the other equation and solve.
  3. Substitute this solution into one of the equations and solve for the value of the other variable. Since the right-hand side of the equation in Equation (I) is simpler, let's use that for our initial substitution. We will substitue this value for y in Equation (II).
    y=-2x-1 & (I) y= 12x-16 & (II)
    y=-2x-1 -2x-1= 12x-16
    â–Ľ
    (II): Solve for x
    y=-2x-1 -2x+15= 12x
    y=-2x-1 - 42x+15= 12x
    y=-2x-1 15= 52x
    y=-2x-1 30=5x
    y=-2x-1 6=x
    y=-2x-1 x=6
    Great! Now, to find the value of y, we need to substitute x=6 into either one of the equations in the given system. Let's use the first equation.
    y=-2x-1 & (I) x=6 & (II)
    y=-2( 6)-1 x=6
    y=-12-1 x=6
    y=-13 x=6
    The solution, or point of intersection, to this system of equations is the point (6, -13).
b Let's take a look at the given system of equations.
y=x^2+1 & (I) y=-2^2 & (II) Just like in Part A, the y variable is already isolated in both equations. Let's use the substitution method to try to solve the system. Since the expression equal to y in Equation (II) is simpler, let's use that for our initial substitution.
y=x^2+1 & (I) y= - x^2 & (II)
- x^2=x^2+1 y= - x^2
0=2x^2+1 y= - x^2
-1=2x^2 y= - x^2
- 12=x^2 y= - x^2
Now, notice that to proceed further, we would need to find the square root of a negative number in Equation (I). However, there is no real number whose square would be negative. For this reason, our system of equation has no real solution. Let's consider the original system of equations. y=x^2+1 y= - x^2 Both of these equations is an equation of a parabola. Since we found that there is no real solution to the system, this means that the two parabolas do not intersect.