Lesson 4.3: Simultaneous Linear Equations

Introduction

In this lesson, students will learn about simultaneous linear equations, which are sets of equations with multiple variables that share a common solution. The primary operations to solve these equations include elimination and substitution methods. By the end of the lesson, students will be able to understand how to solve simultaneous equations and interpret their solutions graphically.

Learning Objectives

• Understand and apply the elimination method to solve a pair of linear equations.
• Understand and apply the substitution method to solve a pair of linear equations.
• Recognize the graphical meaning of the solution as the point of intersection between two lines.
• Solve word problems that result in pairs of simultaneous equations.

Section 1: Solving Simultaneous Equations by Elimination

The elimination method involves manipulating the equations in such a way that one of the variables can be eliminated, allowing for the other variable to be solved easily.

Step-by-step Process of Elimination Method

1. Align the Equations: Write the two equations in a standard form.
2. Eliminate One Variable: Adjust the coefficients of one variable to be equal.
3. Add or Subtract the Equations: This will eliminate one variable, allowing you to solve for the remaining variable.
4. Substitute Back to Find the Other Variable: Once one variable is found, substitute it back into one of the original equations to find the other.

Example 1: Using Elimination

Consider the equations:

$$ 2x + 3y = 12 $$

$$ 4x - 2y = 10 $$

Step 1: Align the equations. They are already in standard form.

Step 2: We want to eliminate $y$. To do this, we can multiply the first equation by $2$:

$$ 4x + 6y = 24 $$

Now, we have:

$$ 4x + 6y = 24 $$

$$ 4x - 2y = 10 $$

Step 3: Subtract the second equation from the first:

$$ (4x + 6y) - (4x - 2y) = 24 - 10 $$

This simplifies to:

$$ 8y = 14 $$

So,

$$ y = \frac{14}{8} = \frac{7}{4} $$

Step 4: Substitute $y = \frac{7}{4}$ back into one of the original equations. Using the first equation:

$$ 2x + 3(\frac{7}{4}) = 12 $$

This becomes:

$$ 2x + \frac{21}{4} = 12 $$

To eliminate the fraction, multiply through by $4$:

$$ 8x + 21 = 48 $$

So,

$$ 8x = 48 - 21 $$

$$ 8x = 27 $$

Thus,

$$ x = \frac{27}{8} $$

Solution:

The solution to the simultaneous equations is:

$$ (x, y) = \left(\frac{27}{8}, \frac{7}{4}

ight) $$

Section 2: Solving Simultaneous Equations by Substitution

In the substitution method, we solve one equation for one variable and substitute that expression into the other equation. This method is particularly useful when one equation is already solved for one variable.

Step-by-step Process of Substitution Method

1. Solve One Equation for One Variable: Isolate one variable (e.g., $y$) in one of the equations.
2. Substitute: Replace the variable in the second equation with the expression found in step 1.
3. Solve for the Remaining Variable: Solve the resulting equation for the remaining variable.
4. Substitute Back: Once found, substitute it back into one of the original equations to find the other variable.

Example 2: Using Substitution

Consider the equations:

$$ y = 2x + 3 $$

$$ 4x + y = 11 $$

Step 1: The first equation is already solved for $y$.

Step 2: Substitute $y$ into the second equation:

$$ 4x + (2x + 3) = 11 $$

Step 3: Combine like terms:

$$ 6x + 3 = 11 $$

Now, subtract $3$ from both sides:

$$ 6x = 8 $$

Thus,

$$ x = \frac{8}{6} = \frac{4}{3} $$

Step 4: Substitute back to find $y$:

$$ y = 2\left(\frac{4}{3}

ight) + 3 $$

This simplifies to:

$$ y = \frac{8}{3} + 3 = \frac{8}{3} + \frac{9}{3} = \frac{17}{3} $$

Solution:

The solution to the simultaneous equations is:

$$ (x, y) = \left(\frac{4}{3}, \frac{17}{3}

ight) $$

Section 3: Graphical Interpretation of Solutions

The geometric representation of simultaneous equations is significant in understanding their solutions graphically. The solutions represent the point of intersection of two lines on the Cartesian plane.

Understanding Graphical Solutions

• Each line represents an equation.
• The intersection point signifies the $x$ and $y$ values that satisfy both equations.
• If the lines intersect at a point, there is one unique solution.
• If they are parallel, they have no solution.
• If they coincide, there are infinitely many solutions.

Example: Graphical Interpretation

Consider the equations:

$$ y = x + 1 $$

$$ y = -x + 5 $$

Graphing both equations, we find:

• The first line has a slope of $1$ (rising) and y-intercept of $1$.
• The second line has a slope of $-1$ (falling) and y-intercept of $5$.

Finding the intersection graphically involves plotting both lines and identifying where they intersect.

From our calculations, the intersection occurs at the point where:

$$ (x, y) = (2, 3) $$

Confirming that by substituting back into both equations will verify the result.

Conclusion

In this lesson, students has explored solving simultaneous linear equations using both the elimination and substitution methods. Understanding these methods provides a solid foundation for dealing with algebraic systems in mathematics and real-world problems. Graphical interpretation allows for a deeper understanding of the solutions, whether they be unique, non-existent, or infinite.

Study Notes

• Simultaneous equations can be solved using either the elimination or the substitution method.
• The elimination method relies on adding or subtracting equations to eliminate a variable.
• The substitution method requires solving one equation for a variable and substituting it into the other equation.
• Graphically, the solution can be interpreted as the point where two lines intersect, indicating where both equations hold true.
• Always check your solutions by substituting back into the original equations.