Solving System Of Equations -3a + 7b = -16 And -9a + 5b = 16
Let's dive into solving this system of linear equations. We are presented with two equations:
- -3a + 7b = -16
- -9a + 5b = 16
There are several methods to tackle this, but we'll focus on the elimination method and the substitution method to provide a comprehensive understanding. Both methods aim to find the values of 'a' and 'b' that satisfy both equations simultaneously.
Method 1: Elimination Method
The elimination method involves manipulating the equations so that when they are added or subtracted, one variable cancels out. To do this, we need to find a common multiple for either the 'a' or 'b' coefficients. Looking at the equations, the 'a' coefficients are -3 and -9. The least common multiple of 3 and 9 is 9. So, we can multiply the first equation by -3 to make the 'a' coefficient 9, which will then allow us to eliminate 'a' when added to the second equation.
Multiply equation 1 by -3:
-3 * (-3a + 7b) = -3 * (-16)
This simplifies to:
9a - 21b = 48
Now we have the modified system:
- 9a - 21b = 48
- -9a + 5b = 16
Next, we add the two equations together. Notice how the '9a' and '-9a' terms cancel each other out:
(9a - 21b) + (-9a + 5b) = 48 + 16
This simplifies to:
-16b = 64
Now, we can solve for 'b' by dividing both sides by -16:
b = 64 / -16
b = -4
Great! We've found the value of 'b'. Now we can substitute this value back into either of the original equations to solve for 'a'. Let's use the first original equation:
-3a + 7b = -16
Substitute b = -4:
-3a + 7(-4) = -16
-3a - 28 = -16
Add 28 to both sides:
-3a = 12
Divide by -3:
a = -4
So, using the elimination method, we find that a = -4 and b = -4.
Method 2: Substitution Method
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. Let's solve the first equation for 'a':
-3a + 7b = -16
Subtract 7b from both sides:
-3a = -16 - 7b
Divide by -3:
a = (16 + 7b) / 3
Now, substitute this expression for 'a' into the second equation:
-9a + 5b = 16
-9 * ((16 + 7b) / 3) + 5b = 16
Simplify:
-3(16 + 7b) + 5b = 16
-48 - 21b + 5b = 16
Combine like terms:
-16b = 64
Divide by -16:
b = -4
We've arrived at the same value for 'b' as before. Now, substitute b = -4 back into the expression we found for 'a':
a = (16 + 7b) / 3
a = (16 + 7(-4)) / 3
a = (16 - 28) / 3
a = -12 / 3
a = -4
Again, we find that a = -4. This confirms our solution using the substitution method.
Conclusion
Therefore, the solution to the system of equations is a = -4 and b = -4. We arrived at this conclusion using both the elimination method and the substitution method, showcasing the robustness of our solution.
Which statement is true about the solution to this system of equations?
The correct statement is:
A. The values of a and b are both equal to -4.
To fully appreciate the solution, let's delve deeper into linear equations and systems of equations. A linear equation is an equation in which the highest power of any variable is 1. This means we won't see terms like a², b³, or square roots of variables. The graph of a linear equation is always a straight line, hence the name “linear.”
A system of linear equations is a set of two or more linear equations that share the same variables. The solution to a system of linear equations is the set of values for the variables that satisfy all equations in the system simultaneously. Graphically, the solution represents the point(s) where the lines corresponding to the equations intersect.
Methods for Solving Systems of Linear Equations
We've already explored two powerful methods, elimination and substitution. However, it’s worth briefly mentioning another method: graphing.
1. Graphing
Each equation in the system represents a line. To solve by graphing, we plot each line on the same coordinate plane. The point(s) where the lines intersect represent the solution(s) to the system. If the lines are parallel, there is no solution. If the lines coincide (are the same line), there are infinitely many solutions.
The graphing method is visually intuitive but can be less precise than algebraic methods like elimination and substitution, especially when the solutions are not integers.
2. Elimination (as discussed above)
The elimination method works by manipulating the equations so that the coefficients of one variable are opposites. This is achieved by multiplying one or both equations by a constant. When the equations are added, one variable is eliminated, allowing us to solve for the remaining variable. The value of the solved variable is then substituted back into one of the original equations to find the value of the other variable.
The elimination method is particularly efficient when the coefficients of one variable are easily made opposites, or when dealing with larger systems of equations.
3. Substitution (as discussed above)
The substitution method involves solving one equation for one variable in terms of the other variable. This expression is then substituted into the other equation, creating a single equation with one variable. Solving this equation yields the value of one variable, which is then substituted back into either original equation (or the solved expression) to find the value of the other variable.
The substitution method is effective when one equation is easily solved for one variable, or when one variable has a coefficient of 1.
Consistent, Inconsistent, and Dependent Systems
Systems of linear equations can be classified based on the number of solutions they have:
- Consistent System: A system that has at least one solution. Consistent systems can be either independent or dependent.
- Independent System: A consistent system that has exactly one solution. The lines intersect at a single point.
- Dependent System: A consistent system that has infinitely many solutions. The equations represent the same line.
- Inconsistent System: A system that has no solution. The lines are parallel and never intersect.
In our example, the system is consistent and independent, as it has a unique solution (a = -4, b = -4).
Applications of Systems of Linear Equations
Systems of linear equations are not just abstract mathematical concepts; they have widespread applications in various fields, including:
- Science and Engineering: Solving for unknowns in physical systems, such as electrical circuits, chemical reactions, and structural analysis.
- Economics: Modeling supply and demand, determining equilibrium prices, and analyzing market behavior.
- Computer Science: Solving linear programming problems, which are used in optimization algorithms and resource allocation.
- Statistics: Regression analysis, where linear equations are used to model relationships between variables.
- Everyday Life: Solving problems involving mixtures, rates, and distances.
For instance, consider a scenario where you're trying to determine the cost of two different items. You know the combined cost of several of each item and another combined cost with different quantities. You can set up a system of two equations with two variables (the cost of each item) and solve for the individual costs.
Conclusion: Mastering Systems of Linear Equations
Solving systems of linear equations is a fundamental skill in mathematics with far-reaching applications. By understanding the underlying concepts and mastering methods like elimination and substitution, you gain a powerful tool for tackling a wide range of problems in diverse fields. Whether you're balancing chemical equations, optimizing a business strategy, or simply solving a real-world puzzle, the ability to work with systems of linear equations is an invaluable asset.
Remember that the key is to practice and become comfortable with different methods and problem-solving approaches. With consistent effort, you can confidently navigate the world of linear equations and their applications.
