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L 3.5.3 Quiz: Nonlinear Systems of Equations

Question 8 of 10

Solve the system of equations:
[tex]
\begin{array}{l}
y = 2x - 5 \\
y = x^2 - 5
\end{array}
[/tex]

A. [tex]\((0, -5)\)[/tex] and [tex]\((2, -1)\)[/tex]

B. [tex]\((0, 5)\)[/tex] and [tex]\((2, 2)\)[/tex]

C. [tex]\((-1, -4)\)[/tex] and [tex]\((3, 4)\)[/tex]

D. [tex]\((-1, -7)\)[/tex] and [tex]\((4, 3)\)[/tex]



Answer :

GinnyAnswer
To solve the system of equations:

1. [tex]\( y = 2x - 5 \)[/tex]
2. [tex]\( y = x^2 - 5 \)[/tex]

we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously.

Let's solve these step-by-step:

### Step 1: Set the equations equal to each other
Since both equations equal [tex]\( y \)[/tex], we can set the right-hand sides of the equations equal to each other:
[tex]\[ 2x - 5 = x^2 - 5 \][/tex]

### Step 2: Simplify the equation
Subtract [tex]\((2x - 5)\)[/tex] from both sides to combine like terms:
[tex]\[ 0 = x^2 - 2x \][/tex]

### Step 3: Factorize the equation
[tex]\[ x^2 - 2x = 0 \][/tex]
[tex]\[ x(x - 2) = 0 \][/tex]

### Step 4: Solve for [tex]\( x \)[/tex]
Set each factor equal to zero:
[tex]\[ x = 0 \][/tex]
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]

Now we have two values for [tex]\( x \)[/tex]: [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex].

### Step 5: Solve for [tex]\( y \)[/tex] using each [tex]\( x \)[/tex] value
Use these [tex]\( x \)[/tex] values to find the corresponding [tex]\( y \)[/tex] values from either of the original equations. We'll use [tex]\( y = 2x - 5 \)[/tex].

- For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0) - 5 = -5 \][/tex]

- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2(2) - 5 = 4 - 5 = -1 \][/tex]

### Step 6: Write the solutions as ordered pairs
The solutions are:
[tex]\[ (0, -5) \text{ and } (2, -1) \][/tex]

So, the correct answer is:
B. [tex]\( (0, -5) \)[/tex] and [tex]\( (2, -1) \)[/tex]

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