Solve the System of Equations y=3x^2-1 y=2x^2+3x+3
The problem asks for a solution to a system of two algebraic equations that represent two different curves, specifically a pair of parabolas in the coordinate plane. The first equation, y = 3x^2 - 1, describes a parabola that opens upwards with a vertex that has been shifted down by 1 unit from the origin. The second equation, y = 2x^2 + 3x + 3, also represents a parabola that opens upwards, but this one is also horizontally shifted and stretched due to the x-term and the constant.
Solving the system involves finding the point(s) of intersection between the two parabolas, which means determining the values of the variable x, and subsequently y, where the two equations have common solutions. This typically involves equating the right-hand sides of the two equations and solving for x to find the intersecting point(s).
$y = 3 x^{2} - 1$$y = 2 x^{2} + 3 x + 3$
Answer
Solution:
Step:1
Set the two equations equal to each other since they both equal $y$.
$$3x^2 - 1 = 2x^2 + 3x + 3$$
Step:2
Isolate $x$ by solving the equation $3x^2 - 1 = 2x^2 + 3x + 3$.
Step:2.1
Rearrange the equation to place $x$ terms on the left side.
$$2x^2 + 3x + 3 = 3x^2 - 1$$
Step:2.2
Consolidate all $x$ terms on one side by subtracting $3x^2$ from both sides.
$$2x^2 + 3x + 3 - 3x^2 = -1$$
Step:2.3
Combine like terms to simplify the equation.
$$-x^2 + 3x + 3 = -1$$
Step:2.4
Balance the equation by adding $1$ to both sides.
$$-x^2 + 3x + 4 = 0$$
Step:2.5
Factor the quadratic equation.
Step:2.5.1
Extract $-1$ from each term on the left.
$$-(x^2 - 3x - 4) = 0$$
Step:2.5.2
Factor the trinomial $x^2 - 3x - 4$.
Step:2.5.2.1
Apply the AC method to factor the quadratic expression.
Find two numbers that multiply to $-4$ and add to $-3$.
Step:2.5.2.2
Write the factors based on the numbers found.
$$-(x - 4)(x + 1) = 0$$
Step:2.6
Set each factor equal to zero to find the values of $x$.
$$x - 4 = 0 \quad \text{or} \quad x + 1 = 0$$
Step:2.7
Solve for $x$ from each factor.
Step:2.7.1
Add $4$ to both sides of the first equation.
$$x = 4$$
Step:2.8
Subtract $1$ from both sides of the second equation.
$$x = -1$$
Step:2.9
The solutions for $x$ are $4$ and $-1$.
Step:3
Calculate $y$ when $x = 4$.
Step:3.1
Plug $x = 4$ into the second equation.
$$y = 2(4)^2 + 3(4) + 3$$
Step:3.2
Simplify to find $y$.
Step:3.2.1
Square $4$ and multiply by $2$.
$$y = 2 \cdot 16 + 3 \cdot 4 + 3$$
Step:3.2.2
Add the results to get $y$.
$$y = 32 + 12 + 3$$ $$y = 47$$
Step:4
Calculate $y$ when $x = -1$.
Step:4.1
Plug $x = -1$ into the second equation.
$$y = 2(-1)^2 + 3(-1) + 3$$
Step:4.2
Simplify to find $y$.
Step:4.2.1
Square $-1$ and multiply by $2$.
$$y = 2 \cdot 1 - 3 + 3$$
Step:4.2.2
Combine the terms to get $y$.
$$y = 2 - 3 + 3$$ $$y = 2$$
Step:5
The solutions to the system are the pairs $(4, 47)$ and $(-1, 2)$.
Step:6
Present the solution in different formats.
Point Form:
$$(4, 47), (-1, 2)$$
Equation Form:
$$x = 4, y = 47$$ $$x = -1, y = 2$$
Knowledge Notes:
To solve a system of equations where both equations are set equal to $y$, you can set the right-hand sides of the equations equal to each other. This is because if $a = b$ and $a = c$, then $b = c$ by the transitive property of equality. The steps involve manipulating the equation to isolate $x$, factoring the resulting quadratic equation, and then solving for $x$. Once $x$ is found, you substitute it back into one of the original equations to find the corresponding $y$ value. The solutions to the system are the pairs of $x$ and $y$ that satisfy both equations simultaneously.
The AC method of factoring is a technique used to factor trinomials of the form $ax^2 + bx + c$. It involves finding two numbers that multiply to $a \cdot c$ and add to $b$. These two numbers are then used to split the middle term and factor by grouping.
When presenting the solution to a system of equations, it can be shown as an ordered pair representing the point of intersection on a graph or in equation form showing the values of $x$ and $y$ that satisfy the system.
