Solve Systems Of Equations A Step-by-Step Guide With Examples
Hey guys! Today, we're diving deep into the fascinating world of solving systems of equations. Specifically, we'll be tackling a system involving both quadratic and linear equations. It might seem intimidating at first, but trust me, with a step-by-step approach, you'll be solving these like a pro in no time! Let's jump right into it!
The System We're Solving
We have the following system of equations:
\begin{cases}
x^2 + y^2 = 7 \\
x^2 - y = 5
\end{cases}
Our goal is to find all the real number pairs (x, y) that satisfy both equations simultaneously. This means that when we plug these values of x and y into both equations, they should hold true. There might be one solution, multiple solutions, or even no solutions at all! That's the exciting part of the puzzle – figuring it all out.
Step 1 The Power of Elimination
The first thing that might catch your eye is that both equations have an x² term. This is a golden opportunity for us to use the elimination method. The elimination method is a technique where we manipulate the equations in a system so that when we add or subtract them, one of the variables disappears. This leaves us with a simpler equation in just one variable, which is much easier to solve.
In our case, since the x² terms have the same coefficient (which is 1), we can simply subtract the second equation from the first equation. This will eliminate the x² term and leave us with an equation involving only y.
Let's do it! Subtracting the second equation (x² - y = 5) from the first equation (x² + y² = 7), we get:
(x² + y²) - (x² - y) = 7 - 5
Simplifying this, we have:
x² + y² - x² + y = 2
The x² terms cancel out, leaving us with:
y² + y = 2
Look at that! We've successfully eliminated x and now have a quadratic equation in terms of y. This is a huge step forward.
Step 2 Taming the Quadratic Equation
Now that we have the quadratic equation y² + y = 2, we need to solve it for y. Remember, a quadratic equation is an equation of the form ay² + by + c = 0, where a, b, and c are constants.
To solve a quadratic equation, we usually try to factor it, use the quadratic formula, or complete the square. In this case, factoring seems like the easiest approach.
First, we need to rearrange the equation so that it's in the standard quadratic form. We can do this by subtracting 2 from both sides:
y² + y - 2 = 0
Now, we need to find two numbers that multiply to -2 and add up to 1 (the coefficient of the y term). Can you think of what they are? They are 2 and -1!
So, we can factor the quadratic equation as follows:
(y + 2)(y - 1) = 0
According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. This means that either y + 2 = 0 or y - 1 = 0.
Solving these two simple equations, we get:
y = -2 or y = 1
Great! We've found two possible values for y. This means we're likely to have two solutions to our system of equations.
Step 3 Finding the Corresponding x Values
Now that we have the y values, we need to find the corresponding x values. To do this, we'll substitute each y value back into one of the original equations. It doesn't matter which equation we choose; we'll get the same x values either way. Let's use the second equation, x² - y = 5, as it seems a bit simpler.
Case 1 When y = -2
Substituting y = -2 into the second equation, we get:
x² - (-2) = 5
Simplifying, we have:
x² + 2 = 5
Subtracting 2 from both sides, we get:
x² = 3
Taking the square root of both sides, we get:
x = ±√3
So, when y = -2, we have two possible x values: x = √3 and x = -√3.
Case 2 When y = 1
Substituting y = 1 into the second equation, we get:
x² - 1 = 5
Adding 1 to both sides, we get:
x² = 6
Taking the square root of both sides, we get:
x = ±√6
So, when y = 1, we have two more possible x values: x = √6 and x = -√6.
Step 4 The Solutions Unveiled
We've done it! We've found all the solutions to the system of equations. Let's list them out as ordered pairs (x, y):
- (√3, -2)
- (-√3, -2)
- (√6, 1)
- (-√6, 1)
These are the four points where the graphs of the two equations intersect. If you were to plot these equations, you'd see that they indeed intersect at these four points.
Verification - The Final Check
It's always a good idea to verify our solutions by plugging them back into the original equations. This helps us catch any errors we might have made along the way.
Let's take the solution (√3, -2) as an example. Plugging these values into the first equation, x² + y² = 7, we get:
(√3)² + (-2)² = 3 + 4 = 7
This checks out! Now, let's plug them into the second equation, x² - y = 5:
(√3)² - (-2) = 3 + 2 = 5
This also checks out! So, the solution (√3, -2) is indeed a valid solution.
You can do the same for the other three solutions to verify that they are also correct.
Conclusion Mastering Systems of Equations
Solving systems of equations can seem daunting at first, but as we've seen, breaking it down into smaller steps makes the process much more manageable. We used the elimination method to simplify the system, solved a quadratic equation by factoring, and then substituted the y values back into the equations to find the corresponding x values. Finally, we verified our solutions to ensure accuracy.
Remember, practice makes perfect! The more you solve these types of problems, the more comfortable you'll become with the techniques involved. Keep practicing, and you'll become a system-solving superstar!
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Repair Input Keyword
Original Question: Solve the following system of equations:
\begin{cases}
x^2 + y^2 = 7 \\
x^2 - y = 5
\end{cases}
If there is more than one solution, enter additional solutions. If there is no real solution, indicate "No solution."
Repaired Question: Find all real solutions (x, y) for the following system of equations:
\begin{cases}
x^2 + y^2 = 7 \\
x^2 - y = 5
\end{cases}
If multiple solutions exist, provide all of them. If no real solutions exist, state "No solution."
SEO Title
Solve Systems of Equations A Step-by-Step Guide with Examples
