Solve the System of Equations y=3x^2-1 y=2x^2+3x+3

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 $x$.

$$x-4=0\text{or}x+1=0$$

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 $a{x}^2+bx+c$. 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.