Solving The System Of Equations Y=x^2+3x-7 And 3x-y=-2
Hey guys! Today, we're diving into the fascinating world of solving systems of equations. Specifically, we'll tackle a system involving a quadratic equation and a linear equation. Don't worry, it's not as scary as it sounds! We'll break it down step by step, making sure you understand the logic behind each move. Our main goal is to find the solutions to the following system:
y = x^2 + 3x - 7
3x - y = -2
And we'll determine which of the following options is the correct solution:
A. (3,11) and ( -3,-7 ) B. (11,3) and ( -3,-7 ) C. (3,11) and ( -7,-3 ) D. No real solutions
So, grab your pencils, and let's get started!
Understanding the Problem
Before we jump into solving, let's make sure we understand what the question is asking. We have two equations. The first one, y = x^2 + 3x - 7, is a quadratic equation, which means its graph is a parabola. The second one, 3x - y = -2, is a linear equation, and its graph is a straight line. The solutions to the system are the points where these two graphs intersect. These points of intersection are represented as ordered pairs (x, y) that satisfy both equations simultaneously.
Essentially, we're looking for the x and y values that make both equations true at the same time. There are a couple of ways we can approach this. One method is substitution, where we solve one equation for one variable and then substitute that expression into the other equation. Another method is elimination, where we manipulate the equations to eliminate one variable. In this case, substitution seems like the more straightforward approach, so let's go with that.
To effectively use the substitution method, we need to identify an equation that is easily solved for one of the variables. Looking at our system, the second equation, 3x - y = -2, seems like a good candidate because we can easily isolate 'y'. Isolating a variable allows us to express it in terms of the other, which is crucial for substitution. So, the first strategic step we'll take is rearranging 3x - y = -2 to solve for 'y'. By doing so, we will get an expression for 'y' that we can then plug into the first equation. This approach will help us reduce the system of two equations with two variables into a single equation with only one variable, which is much easier to solve. This is the essence of the substitution method: simplifying the problem by reducing the number of variables we need to handle simultaneously.
Solving by Substitution
Our first step is to solve the second equation, 3x - y = -2, for y. To do this, we can add y to both sides and add 2 to both sides, which gives us:
3x + 2 = y
Now we have y isolated. The next crucial step is to substitute this expression for y (which is 3x + 2) into the first equation, y = x^2 + 3x - 7. This substitution will replace the 'y' in the first equation with '3x + 2', allowing us to create a new equation that only involves 'x'. By doing this, we eliminate 'y' as a variable, and we are left with a single equation in terms of 'x' that we can solve. This is the power of the substitution method: it simplifies the system by reducing it to a single equation with a single unknown. The resulting equation will be a quadratic equation, which we can then solve using various methods such as factoring, completing the square, or the quadratic formula. So, let's go ahead and make that substitution and see what we get.
Substituting y = 3x + 2 into the first equation, we get:
3x + 2 = x^2 + 3x - 7
Now we have a single equation with only x as the variable. This is a quadratic equation, and to solve it, we need to get it into the standard form, which is ax^2 + bx + c = 0. Let's rearrange the equation by subtracting 3x from both sides and subtracting 2 from both sides. This process of rearrangement is essential for applying standard methods to solve quadratic equations, such as factoring or using the quadratic formula. Having the equation in standard form allows us to easily identify the coefficients a, b, and c, which are necessary for these methods. By bringing all terms to one side and setting the equation equal to zero, we create a format that is conducive to solving for the roots or solutions of the quadratic equation.
Doing so, we get:
0 = x^2 - 9
Solving the Quadratic Equation
We now have the quadratic equation x^2 - 9 = 0. There are a couple of ways we can solve this. One way is to recognize that this is a difference of squares, which can be factored. Another way is to simply add 9 to both sides and then take the square root. Let's use the difference of squares method first. The difference of squares pattern is a^2 - b^2 = (a + b)(a - b). In our case, a is x and b is 3 (since 9 is 3 squared). This powerful factoring technique allows us to quickly break down certain quadratic expressions into a product of two binomials. Recognizing and applying this pattern can often save time and effort compared to using other methods like the quadratic formula. The difference of squares is a fundamental concept in algebra, and mastering it is essential for solving various types of equations and simplifying expressions. So, by applying this pattern, we can easily factor our quadratic equation and find the values of x that make the equation true.
Factoring x^2 - 9, we get:
(x + 3)(x - 3) = 0
Now, for the product of two factors to be zero, at least one of them must be zero. So, we set each factor equal to zero and solve for x:
x + 3 = 0 or x - 3 = 0
Solving these equations, we get:
x = -3 or x = 3
These are our x-values. Now we need to find the corresponding y-values. To do this, we substitute each x-value back into either of the original equations. Since we already have y = 3x + 2, let's use that one. This step of back-substitution is crucial to find the complete solutions to the system of equations. We have found the x-values where the two equations intersect, but we also need the corresponding y-values to define the points of intersection. By plugging the x-values back into one of the original equations (or the simplified form we derived during substitution), we can calculate the y-values. It's important to use the same equation for all substitutions to ensure consistency and accuracy. This process gives us the ordered pairs (x, y) that represent the solutions to the system.
Finding the Corresponding y-values
Let's start with x = 3. Substituting this into y = 3x + 2, we get:
y = 3(3) + 2 = 9 + 2 = 11
So, when x = 3, y = 11. This gives us the solution (3, 11). Now let's do the same for x = -3:
y = 3(-3) + 2 = -9 + 2 = -7
So, when x = -3, y = -7. This gives us the solution (-3, -7). We now have two solutions: (3, 11) and (-3, -7). It's a good practice to always check these solutions by plugging them back into the original equations to make sure they satisfy both. This verification step helps to catch any potential errors made during the solving process. By substituting the x and y values of each solution into both equations, we can confirm that the equations hold true. If a solution does not satisfy both equations, it indicates a mistake in our calculations, and we need to go back and review our steps. This final check ensures that our solutions are accurate and reliable.
Checking the Solutions
Let's check the solution (3, 11) in both equations:
Equation 1: y = x^2 + 3x - 7
11 = (3)^2 + 3(3) - 7
11 = 9 + 9 - 7
11 = 11 (True)
Equation 2: 3x - y = -2
3(3) - 11 = -2
9 - 11 = -2
-2 = -2 (True)
The solution (3, 11) works! Now let's check (-3, -7):
Equation 1: y = x^2 + 3x - 7
-7 = (-3)^2 + 3(-3) - 7
-7 = 9 - 9 - 7
-7 = -7 (True)
Equation 2: 3x - y = -2
3(-3) - (-7) = -2
-9 + 7 = -2
-2 = -2 (True)
The solution (-3, -7) also works! So, our solutions are (3, 11) and (-3, -7).
The Answer
Looking back at the options, we see that the correct answer is:
A. (3,11) and ( -3,-7 )
We did it! We successfully solved the system of equations using the substitution method. Remember guys, practice makes perfect. The more you work through these types of problems, the more comfortable you'll become with them. Keep up the great work, and I'll see you in the next one!
