How to Solve Systems of Linear and Quadratic Equations with Examples
Linear and quadratic equations are the algebraic systems of equations consisting of one linear equation and one quadratic equation. The goal of solving systems linear quadratic equations is to significantly reduce two equations having two variables down to a single equation with only one variable. Since each equation in the system consists of two variables, one way to decrease the number of variables in an equation is by substituting an expression for a variable.
In a test, you will be expected to identify the solution(s) to systems either algebraically or graphically.
Examples of Linear Equation and Quadratic Equation
Following are the example of Linear and quadratic equations:
y = x + 1
y = x² −1
In systems linear quadratic equations, both equations are simultaneously true. Simply to say, because the 1st equation tells us that y is equivalent to x + 1, the y in the 2nd equation is also equivalent to x + 1. Hence, we can plug in x + 1 as a substitute for y in the 2nd equation:
y = x² − 1
x + 1 = x² −1
From here, we are able to solve and simplify the quadratic equation for x, which provides us with the x-values of the solutions to the linear-quadratic system. Then, we can also use the x-values and either equation in the system in order to find out the y-values.
Solving Linear-Quadratic Systems
You might have solved systems of linear equations. But what about a system of 2 equations where 1 equation is linear and the remaining is quadratic?
We can apply a version of the substitution technique in order to solve systems of this type.
Note that the standard form of the equation for a parabola with a vertical axis of symmetry is y = ax² + bx + c, a ≠ 0 and the slope-intercept form of the equation for a line is y = mx + b,
To avoid any sort of confusion with the variables, write the linear equation as y= mx+d where,
m = The slope.
d = The y-intercept of the line.
Substitute the expression for y from the linear equation, into the quadratic equation. In other words, substitute mx + d for y in y = ax² + bx + c.
mx + d = ax² + bx + c
Now, rewrite the new quadratic equation in the standard form.
Subtract mx+d from both sides of the equation
(mx + d) − (mx + d) = (ax² + bx + c) − (mx + d)0 = ax² + (b − m)x + (c − d)
Now we have a quadratic equation in 1 variable, the solution of which can be determined with the help of the quadratic formula.
The solutions to the equation ax² + (b − m)x + (c − d) = 0 will provide us with the x-coordinates of the points of intersection of the graphs of the parabola and the line. The corresponding y-coordinates can be identified with the help of the linear equation.
Another way of solving the system is to graph the two functions on a similar coordinate plane and then determine the points of intersection.
Solved Examples
Example:
Determine the points of bisection between the line y = 2x + 1 and the parabola y = x² − 2.
Solution:
Substitute the values in the equation 2x + 1 for y in y = x² − 2.
2x + 1 = x² − 2
Now, express the quadratic equation in standard form.
2x + 1− 2x − 1 = x² − 2 − 2x − 10 = x² − 2x − 3
Apply the quadratic formula in order to identify the roots of the quadratic equation.
Here,
a = 1
b = −2
c = −3
x = √[− (−2) ± (−2)2 − 4(1)(−3)]/2(1)
= √[2 ± 4 + 12]/2
= 2 ± 4/2
= 3, −1
Then, Substitute the x-values in the linear equation in order to identify the corresponding y-values.
x = 3 ⇒ y = 2(3) + 1 = 7
x = −1 ⇒ y = 2(−1) + 1 = −1
Thus, the points of bisection are (3, 7) and (−1, −1).
Also, remember that the same method can be used to determine the intersection points of a line and a circle.
Check below the Graphing of the parabola and the straight line on a coordinate plane.
Example: Determine the points of intersection between the line y = −3x and the circle x² + y² = 3.
Solution:
Substitute −3x for y in x² + y² = 3.
x² + (−3x)² = 3
Simplify the system of quadratic equations
x² + 9x² = 3
10x² = 3
x² = 3/10
Taking square roots, x = ±√(3/10).
Now, Substituting the x-values in the linear equation in order to identify the corresponding y-values.
x = √(3/10) ⇒ y = −3(√(3/10))
=−(3/3√10)
x = −√(3/10) ⇒ y =−3(−√(3/10))
= 3/3√10
Hence, the points of intersection come out to be (√(3/10), (-3√3/10), and (−3/√10√, 3√3/√10).
Refer below for the Graphing of the circle and the straight line on a coordinate plane.
Fun Facts
While solving quadratic simultaneous equations, if we get a negative number as the square of a number in the outcome, then the two equations do not have real solutions.
A quadratic and linear system can be represented by a line and a parabola in the xy-plane.
Each intersection of the line and the parabola depicts a solution to a linear-quadratic system.
FAQs on Systems of Linear and Quadratic Equations Explained
1. What exactly is a system of linear and quadratic equations?
A system of linear and quadratic equations is a set of two equations that you solve together. One equation is linear, which forms a straight line when graphed, and the other is quadratic, which forms a curve called a parabola. The solution to the system is the point or points where the line and the parabola intersect.
2. How can you find the solution to one of these systems just by looking at its graph?
The solution is simply the point where the graphs of the two equations cross. If you draw the straight line (from the linear equation) and the parabola (from the quadratic equation) on the same coordinate plane, the points of intersection are the solutions. This visual method helps you see how many solutions are possible.
3. Why can a system of linear and quadratic equations have two, one, or even no solutions?
The number of solutions depends entirely on how the line and the parabola interact on a graph.
- Two solutions: This happens when the line cuts through the parabola at two distinct points.
- One solution: This occurs when the line is tangent to the parabola, meaning it touches the curve at exactly one point.
- No solution: This happens when the line and the parabola never touch or cross each other at all.
4. What is the main algebraic method used to solve these systems?
The most common and straightforward method is substitution. First, you rearrange the linear equation to isolate one variable (like 'y'). Then, you substitute this expression into the quadratic equation. This creates a new equation with only one variable, which you can then solve to find its value.
5. Where might you see systems of linear and quadratic equations used in real life?
These systems are useful for modelling situations where a straight-line path interacts with a curved one. For instance, in physics, they can determine when a thrown object (following a parabolic path) will hit a sloped ground (a straight line). In business, they can find the break-even points between linear costs and quadratic revenue.
6. How is solving a linear-quadratic system different from solving a system of two linear equations?
When solving a system with two linear equations, you're finding where two straight lines meet, which is usually just one point. But with a linear-quadratic system, you're finding where a straight line intersects a curve. Because a line can cut a parabola twice, you often have to find two possible solutions, and the algebra involves solving a quadratic equation rather than a simpler linear one.
7. What is a common mistake to avoid when solving these systems algebraically?
A frequent error is stopping after you find the x-values. A complete solution is an ordered pair (x, y) that satisfies both equations. After you solve the quadratic equation for 'x', you must substitute those x-values back into the original linear equation to find the corresponding y-values for each solution.
