System of Equations
- Introduction
- Understanding Systems of Equations
- Types of Solutions for System of Equations
- Methods for Solving System of Equations
- Solving System of Equations Using Substitution Method
- Solving System of Equations Using Elimination Method
- Solving System of Equations by Graphing
- Solving System of Equations Using Cross-Multiplication Method
- Solving System of Equations Using Matrices
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
A system of equations, also known as a set of simultaneous equations, is a collection of equations with the aim of finding common solutions. These equations typically involve multiple variables and provide relationships between them. The goal is to find values for the variables that satisfy all equations in the system simultaneously.
Solving systems of equations often requires employing different methods, such as substitution, elimination, or graphing. Each method offers its own advantages and is suited for particular types of problems. For instance, substitution involves replacing one variable with an expression in terms of another variable, simplifying the system into a single equation. Elimination involves manipulating equations to eliminate one variable, reducing the system to a simpler form. Graphing involves plotting the equations on a graph to visually identify the intersection points, which represent the solutions to the system.
Understanding Systems of Equations
In algebra, a system of equations means having two or more equations that we're trying to solve together to find common solutions. Specifically, when we talk about linear equations, it's about finding values for the variables that satisfy all the equations in the system.
Example of a System of Equations
`1`. \(3x + 2y = 18\)
`2`. \(2x - 4y = -4\)
Solving the above system of equations means looking for values of \(x\) and \(y\) that satisfy both equations simultaneously.
Solving the System of Equations
To solve this system, we need to find values for \(x\) and \(y\) that make both equations true at the same time. We can use methods like substitution or elimination to determine the solutions. It's like solving a puzzle – once we find the right pieces (values of \(x\) and \(y\)), everything falls into place, and the equations make sense together.
Types of Solutions for System of Equations
A system of equations can have different types of solutions, depending on how the equations interact with each other.
Unique Solution
A system of equations has a unique solution when there's only one set of variables that works for all the equations. Visually, this means the lines representing the equations intersect at a single point on a graph.
No Solution
Sometimes, a system of equations has no solution. This occurs when there's no set of variables that makes all the equations true. Graphically, if we plot these equations, the lines will be parallel, never intersecting.
Infinitely Many Solutions
In certain cases, a system of equations can have infinitely many solutions. This happens when there are countless sets of variables that satisfy all the equations. Graphically, the lines representing the equations overlap perfectly, essentially representing the same line. Since overlapping lines have infinite points each point becomes a solution of the system.
Methods for Solving System of Equations
There are various ways to solve a system of equations. Every method has its unique benefits and is appropriate for specific types of problems. The most commonly used methods are listed below.
- Substitution Method
- Elimination Method
- Graphical Method
- Cross-Multiplication Method
There are usually `2` or more equations required to solve a system of equations in `2` variables. In the same way, we would need at least `3` equations for systems with `3` variables. Let's now explore the three methods for resolving a `2`-variable system of linear equations.
Solving System of Equations Using Substitution Method
When we solve a system of equations using the substitution method, we aim to express one variable in terms of the other and then substitute this expression into the second equation. Let's illustrate this process with an example.
Example: Solve the following system of linear equations using the substitution method.
\(3x - y = 6\)
\(2x + y = 10\)
Solution:
Let’s label the two equations as `(1)` and `(2)`. This will make it easier for us to refer them in the subsequent steps.
\(3x - y = 6\) `→ (1)`
\(2x + y = 10\) `→ (2)`
First, from equation `(1)`, let's express \(y\) in terms of \(x\):
\(y = 3x - 6\) `→ (3)`
Now, substitute this expression for \(y\) into equation `(2)`:
\(2x + (3x - 6) = 10\)
This simplifies to:
\(2x + 3x - 6 = 10\)
Combine like terms:
\(5x - 6 = 10\)
Add `6` to both sides:
\(5x = 16\)
Now, solve for \(x\):
\(x = \frac{16}{5}\)
Substitute this value of \(x\) back into equation `(1)` to find \(y\):
\(3(\frac{16}{5}) - y = 6\)
Solve for \(y\):
\(y = \frac{18}{5}\)
Thus, the solution to the system of equations is \(x = \frac{16}{5}\) and \(y = \frac{18}{5}\).
Solving System of Equations Using Elimination Method
The elimination method is a technique for solving a system of equations by manipulating the equations to eliminate one of the variables. This is done by multiplying the equations by suitable numbers so that the coefficients of one variable become the opposite and cancel out. This leaves us with one variable equation which can be solved to get value for one variable. This value can then be used to get the value for the second variable. This process is repeated till we find the value of all the variables.
Example: Solve the following system of linear equations using the elimination method.
\(3x + 4y = 14\) `→ (1)`
\(2x - 3y = 7\) `→ (2)`
Solution:
Let’s label the two equations as `(1)` and `(2)`. This will make it easier for us to refer them in the subsequent steps.
\(3x + 4y = 14\) `→ (1)`
\(2x - 3y = 7\) `→ (2)`
To eliminate the variable \(y\), we can multiply equation `(1)` by `3` and equation `(2)` by `4`. This makes the coefficients of \(y\) equal and thus easily eliminated.
This yields:
\(9x + 12y = 42\) `→ (3)`
\(8x - 12y = 28\) `→ (4)`
Now, if we add equation `(4)` to equation `(3)`, the \(y\) terms cancel out:
\((9x + 12y) + (8x - 12y) = 42 + 28\)
\(9x + 12y + 8x - 12y = 42 + 28\)
\(17x = 70\)
Now, let's solve for \(x\):
\(x = \frac{70}{17}\)
Next, we can plug the value of \(x\) into either equation `(1)` or `(2)` to find \(y\).
Let's use equation `(1)` in this case:
\(3(\frac{70}{17}) + 4y = 14\)
\(\frac{210}{17} + 4y = 14\)
\(4y = \frac{28}{17}\)
\(y = \frac{28}{68}\)
\(y = \frac{7}{16}\)
Therefore, the solution to the system of equations is \(x = \frac{70}{17}\) and \(y = \frac{7}{16}\).
Solving System of Equations by Graphing
When we use the graphing method to solve a system of linear equations, we plot their graphs on a coordinate plane and find the point where the lines intersect. This intersection point represents the solution to the system of equations. The `x`-coordinate and the `y`-coordinate of the intersection point represent the `x` and `y` values that satisfy the system of equations.
Example: Solve the system of linear equations by graphing method.
`1`. \(3x - y = 7\)
`2`. \(x - 2y = 4\)
Solution:
Plot the graphs of the two equations on a coordinate grid.
The straight lines representing `(3x - y = 7)` and `(x - 2y = 4)` intersect at `(2,-1)`. The point `(2,-1)` lies on both the lines, meaning the `x` and `y` value of the point satisfies both the equations.
Hence, the solution to the given system of equations is `x = 2` and `y = -1`.
Solving System of Equations Using Cross-Multiplication Method
The cross-multiplication method is a technique used to solve systems of equations by equating the ratios of corresponding terms. This method is particularly useful when dealing with equations involving proportions or ratios.
Example: Solve the system of linear equations using the cross-multiplication method.
`1`. \(4x + 2y = 10\)
`2`. \(x - y = 13\)
Solution:
To apply the cross-multiplication method, we'll isolate one variable in each equation and then equate the ratios of corresponding terms.
From equation `1`:
\(4x = 10 - 2y\)
\(x = \frac{10 - 2y}{4}\)
From equation `2`:
\(x - y = 13\)
\(x = y + 13\)
Now, we'll equate the two expressions for \(x\):
\(\frac{10 - 2y}{4} = y + 13\)
Next, we'll cross-multiply:
\( 10 - 2y = 4(y + 13)\)
Expand and solve for \(y\):
\(10 - 2y = 4y + 52\)
\(-2y - 4y = 52 - 10\)
\(-6y = 42\)
\(y = \frac{42}{-6}\)
\(y = -7\)
Now that we have found the value of \(y\), we'll substitute it back into one of the original equations to find \(x\). Let's use equation `2`:
\(x - (-7) = 13\)
Solve for \(x\):
\(x + 7 = 13\)
\(x = 6\)
Therefore, the solution to the system of equations is \(x = 6\) and \(y = -7\).
Solving System of Equations Using Matrices
This method of solving a system of equations using matrices is mostly preferable when we have `3` or more equations in the system. To do this, we express the given equations in standard form, with the variables and constants on their respective sides. Let us illustrate this through an example:
Example: Solve the system of equations using matrices.
\(-3x - 2y + 4z = 9\)
\(3y - 2z = 5\)
\(4x - 3y + 2z = 7\)
Solution:
\(-3x - 2y + 4z = 9\)
\(3y - 2z = 5\)
\(4x - 3y + 2z = 7\)
We can represent them in matrix form as:
\( \begin{bmatrix} -3 & -2 & 4 \\ 0 & 3 & -2 \\ 4 & -3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix} \)
Here,
\( A = \begin{bmatrix} -3 & -2 & 4 \\ 0 & 3 & -2 \\ 4 & -3 & 2 \end{bmatrix} \),
\( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \),
and
\( B = \begin{bmatrix} 9 \\ 5 \\7 \end{bmatrix} \)
The solution of the system is given by the formula \( X = A^{-1}B \), where \( A^{-1} \) denotes the inverse of matrix \( A \). This inverse matrix \( A^{-1} \) can be found using matrix operations.
\( A^{-1} = \begin{bmatrix} 0 & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{11}{16} & \frac{3}{16} \\ \frac{3}{8} & \frac{17}{32} & \frac{9}{32} \end{bmatrix} \)
\( A^{-1}B = \begin{bmatrix} 3 \\ 7 \\ 8 \end{bmatrix} \)
Since \( X = A^{-1}B \), we can write
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \\ 8 \end{bmatrix}\)
Hence \( x = 3 \), \( y = 7 \) and \( z = 8 \).
Solved Examples
Example `1`. Solve the system of equations using the substitution method:
\(3x + 2y = 14\)
\(4x - y = 8\)
Solution:
From the second equation, isolate \(y\):
\(y = 4x - 8\)
Substitute this expression for \(y\) into the first equation:
\(3x + 2(4x - 8) = 14\)
\(3x + 8x - 16 = 14\)
\(11x - 16 = 14\)
\(11x = 30\)
\(x = \frac{30}{11}\)
Now, substitute \(x\) back into either equation to find \(y\):
\(4\left(\frac{30}{11}\right) - y = 8\)
\(y = 4\left(\frac{30}{11}\right) - 8\)
\(y = \frac{120}{11} - \frac{88}{11}\)
\(y = \frac{32}{11}\)
Therefore, the solution is \(x = \frac{30}{11}\) and \(y = \frac{32}{11}\).
Example `2`. Solve the system of equations using the elimination method:
\(2x - 3y = 11\)
\(4x + 5y = 17\)
Solution:
Multiply the first equation by `4` and the second equation by `2` to eliminate \(x\):
\(8x - 12y = 44\)
\(8x + 10y = 34\)
Subtract the second equation from the first:
\((-12y - 10y) = (44 - 34)\)
\(-22y = 10\)
\(y = -\frac{5}{11}\)
Substitute \(y\) back into one of the original equations to find \(x\):
\(2x - 3(-\frac{5}{11}) = 11\)
\(2x + \frac{15}{11} = 11\)
\(2x = 11 - \frac{15}{11}\)
\(2x = \frac{121 - 15}{11}\)
\(2x = \frac{106}{11}\)
\(x = \frac{53}{11}\)
The solution is \(x = \frac{53}{11}\) and \(y = -\frac{5}{11}\).
Example `3`. Solve the system of equations by graphing:
\(4x + y = 2\)
\(x - y = 3\)
Solution:
Plot the graphs of the two equations and find the point of intersection, which is the solution. The intersection point is at `(1, -2)`.
Therefore, the solution is \(x = 1\) and \(y = -2\).
Example `4`. Solve the system of equations using the cross-multiplication method:
\(2x + 3y = 11\)
\(4x - 2y = 10\)
Solution:
From the first equation, we can isolate \(x\):
\(x = \frac{11 - 3y}{2}\)
From the second equation, we can isolate \(x\) as well:
\(x = \frac{10 + 2y}{4}\)
Equating these expressions, we can solve for \(y\) and then find \(x\).
After solving, we get \(x = \frac{13}{4}\) and \(y = \frac{3}{2}\).
Therefore, the solution is \(x = \frac{13}{4}\) and \(y = \frac{3}{2}\).
Practice Problems
Q`1`. Solve the system of equations:
\(x - y = 3\)
\(7x - y = -3\)
- \(x = 1\), \(y = 4\)
- \(x = -1\), \(y = -4\)
- \(x = 2\), \(y = -3\)
- \(x = -2\), \(y = -3\)
Answer: b
Q`2`. Solve the system of equations:
\(3x - 2y = 9\)
\(5x + 4y = 26\)
- \(x = 4\), \(y = \frac{3}{2}\)
- \(x = -5\), \(y = \frac{3}{2}\)
- \(x = \frac{5}{2}\), \(y = -4\)
- \(x = \frac{3}{2}\), \(y = 2\)
Answer: a
Q`3`. Solve the system of equations:
\(x + y = 8\)
\(3x - 2y = 9\)
- \(x = 3\), \(y = 5\)
- \(x = 4\), \(y = 4\)
- \(x = 5\), \(y = 3\)
- \(x = 6\), \(y = 2\)
Answer: c
Q`4`. Solve the system of equations:
\(2x + y = 10\)
\(4x - 3y = 10\)
- \(x = 2\), \(y = 4\)
- \(x = 3\), \(y = 3\)
- \(x = 4\), \(y = 2\)
- \(x = 5\), \(y = 1\)
Answer: c
Q`5`. Solve the system of equations:
\(3x - y = 5\)
\(2x + 2y = 14\)
- \(x = 2\), \(y = 1\)
- \(x = 3\), \(y = 4\)
- \(x = 4\), \(y = 3\)
- \(x = 5\), \(y = 4\)
Answer: b
Q`6`. Solve the system of equations:
\(x - 3y + z = 8\)
\(2x - 5y - 3z = 2\)
\(x + 4y + z = 2\)
- \(x = \frac{12}{5}, \quad y = -1, \quad z = \frac{13}{5} \)
- \(x = \frac{2}{5}, \quad y = 1, \quad z = -\frac{3}{5} \)
- \(x = 4\), \(y = 3\), \(z = 5\)
- \(x = 5\), \(y = 4\), \(z = 3\)
Answer: a
Frequently Asked Questions
Q`1`. What are systems of equations, and why are they important?
Answer: A system of equations are set of two or more equations with the same variables. They are important because they represent real-world situations where multiple variables interact, such as in engineering, economics, and physics. Solving systems of equations helps in finding the values of variables that satisfy all equations simultaneously, providing solutions to complex problems.
Q`2`. What are the common methods for solving systems of equations?
Answer: The common methods for solving systems of equations include:
- Substitution Method
- Elimination Method
- Graphical Method
- Matrix Method
- Cross-Multiplication Method
Q`3`. How do I know which method to use when solving a system of equations?
Answer: The choice of method depends on the specific equations and variables involved. Here are some guidelines:
- Use the Substitution Method when one equation can be easily solved for one variable.
- Use the Elimination Method when adding or subtracting equations will eliminate one of the variables.
- Use the Graphical Method when dealing with simple equations and when visualizing the solutions is helpful.
- Use the Matrix Method when dealing with larger systems of equations, like three or more variables.
- Use the Cross-Multiplication Method when solving systems with linear equations and multiple variables.
Q`4`. Can systems of equations have more than one solution?
Answer: Yes, systems of equations can have:
- A unique solution: One set of values for the variables that satisfy all equations.
- No solution: When the equations represent parallel lines that never intersect.
- Infinitely many solutions: When the equations represent the same line, resulting in infinitely many points of intersection.
Q`5`. What are some real-world applications of solving systems of equations?
Answer: Solving systems of equations is widely used in various fields, where many parameters are taken into account simultaneously. For example:
- Engineering: Designing structures, circuits, and systems.
- Economics: Analyzing supply and demand, cost and revenue functions.
- Physics: Solving problems involving motion, forces, and energy.
- Biology: Modeling population growth, and biochemical reactions.
- Chemistry: Balancing chemical equations, and calculating reaction rates.