4.3 Systems of Linear Equations and Inequalities in Two Variables

Systems of Linear Equations and Inequalities in Two Variables

When we deal with situations involving two unknown quantities, we often need two distinct equations to find a unique solution. A set of two or more equations sharing the same variables is called a system of equations. The solution to a system of linear equations is the ordered pair $(x, y)$ that makes all equations in the system true simultaneously.

Methods for Solving Systems of Linear Equations

There are three primary methods for solving systems of linear equations: graphing, substitution, and elimination.

1. Solving by Graphing

To solve by graphing, you plot both linear equations on the same coordinate plane. The solution is the exact point $(x, y)$ where the two lines intersect. While visually intuitive and helpful for understanding the concept of a solution, graphing can be imprecise if the intersection point involves fractions or decimals.

2. Solving by Substitution

The substitution method is highly effective when one of the equations is already solved for a specific variable, or if one variable has a coefficient of $1$ or $-1$.

1. Solve one of the equations for one variable in terms of the other.
2. Substitute that entire expression into the other equation. You now have an equation with only one variable.
3. Solve the new equation for that single variable.
4. Substitute the resulting value back into your rearranged first equation to find the value of the second variable.

Example: Solve $y = 2x + 1$ and $3x + y = 11$. Since $y$ is already isolated, substitute $(2x + 1)$ into the second equation: $3x + (2x + 1) = 11$. Combine terms: $5x + 1 = 11 \rightarrow 5x = 10 \rightarrow x = 2$. Substitute $x = 2$ back into the first equation: $y = 2(2) + 1 \rightarrow y = 5$. The solution is $(2, 5)$.

3. Solving by Elimination (Addition/Subtraction Method)

The elimination method involves adding or subtracting the equations to eliminate one of the variables entirely. This is usually the most efficient method when both equations are written in standard form ($Ax + By = C$).

1. If necessary, multiply one or both equations by a constant so that the coefficients of one of the variables are opposites (e.g., $3x$ and $-3x$).
2. Add the two equations together. The chosen variable will cancel out.
3. Solve for the remaining variable.
4. Substitute that value back into either original equation to solve for the eliminated variable.

System Classification

Not all systems have exactly one solution. Systems are classified based on the number of solutions they possess:

• Independent and Consistent: The lines have different slopes and intersect at exactly one point. There is one solution.
• Dependent and Consistent: The equations represent the exact same line. They have the same slope and same y-intercept. Any point on the line is a solution, so there are infinitely many solutions.
• Inconsistent: The equations represent parallel lines. They have the same slope but different y-intercepts. Since they never intersect, there is no solution.

Word Problem Applications

Systems of equations are incredibly useful for solving real-world word problems. Common types include:

• Mixture Problems: Combining two different items of known costs/concentrations to create a specific final mixture. (e.g., blending cheap and expensive coffee beans).
• Motion Problems: Involving speed, distance, and time, often with an added variable like a headwind or water current affecting the travel.
• Cost/Revenue Problems: Finding the break-even point where the cost to produce items equals the revenue generated from selling them.

Systems of Linear Inequalities

A system of linear inequalities consists of two or more inequalities. The solution is not a single point, but rather a region on the coordinate plane. To solve, graph each inequality on the same plane:

1. Graph the boundary line. Use a solid line for $\le$ or $\ge$, and a dashed line for $<$ or $>$.
2. Shade the appropriate side of the line using a test point (like $(0,0)$).
3. The final solution to the system is the overlapping shaded region where all individual shadings intersect. Any point in this overlapping area satisfies all inequalities in the system.

Detailed Worked Example: The Substitution Method

Let's consider a practical scenario where the substitution method shines. Suppose we have the system: $y = 4x - 2$ $2x + 3y = 22$

Step 1: Since $y$ is already isolated in the first equation, we substitute $(4x - 2)$ for $y$ in the second equation: $2x + 3(4x - 2) = 22$

Step 2: Distribute the $3$: $2x + 12x - 6 = 22$

Step 3: Combine like terms: $14x - 6 = 22$

Step 4: Add $6$ to both sides: $14x = 28$

Step 5: Divide by $14$: $x = 2$

Step 6: Substitute $x = 2$ back into the first equation: $y = 4(2) - 2$ $y = 8 - 2 = 6$ The solution is $(2, 6)$.

Teaching Scenario: Analyzing Student Errors in Elimination

A common error students make when using the elimination method involves incorrectly distributing negative signs when subtracting equations. For instance, consider the system: $5x + 2y = 15$ $5x - 3y = 5$ A student might subtract the second equation from the first to eliminate $x$, but erroneously compute $2y - (-3y)$ as $-1y$. The correct computation is $2y + 3y = 5y$, leading to $5y = 10$, and thus $y = 2$. Teachers must emphasize that subtracting a negative is equivalent to adding a positive. Encouraging students to write out the subtraction step explicitly or to multiply the entire second equation by $-1$ and then add can significantly reduce these errors.

Step-by-Step Derivation of a System's Solution

Let's derive the algebraic solution for a general system of two linear equations: $ax + by = c$ $dx + ey = f$

To eliminate $y$, we can multiply the first equation by $e$ and the second by $b$: $aex + bey = ce$ $bdx + bey = bf$

Subtracting the second from the first yields: $(ae - bd)x = ce - bf$ $x = \frac{ce - bf}{ae - bd}$ (assuming $ae - bd \neq 0$)

Similarly, to eliminate $x$, multiply the first by $d$ and the second by $a$: $adx + bdy = cd$ $adx + aey = af$

Subtracting the second from the first gives: $(bd - ae)y = cd - af$ $y = \frac{cd - af}{bd - ae} = \frac{af - cd}{ae - bd}$

This derivation shows the roots of Cramer's Rule, which uses determinants to solve systems!

Further Application: The Motion Problem

Consider a boat traveling downstream and upstream. The boat travels 36 miles downstream in 2 hours. The return trip upstream takes 3 hours. What is the speed of the boat in still water, and what is the speed of the current? Let $b$ be the boat's speed and $c$ be the current's speed. Downstream speed: $b + c$ Upstream speed: $b - c$ Using Distance = Rate $\times$ Time: $36 = 2(b + c) \Rightarrow b + c = 18$ $36 = 3(b - c) \Rightarrow b - c = 12$ Adding the two simplified equations: $2b = 30 \Rightarrow b = 15$ Substituting $b = 15$ into $b + c = 18$: $15 + c = 18 \Rightarrow c = 3$ The boat's speed is 15 mph, and the current is 3 mph.

Geometric Interpretation of Dependent Systems

When a system is dependent, the two equations describe the same line. For example, $2x + y = 4$ and $4x + 2y = 8$. Dividing the second equation by 2 yields the first. This means any coordinate pair $(x,y)$ that satisfies the first equation also satisfies the second. There is an infinite set of solutions, forming the entire line.

Solving Non-Linear Systems (Linear + Quadratic)

ETS explicitly assesses systems that mix a linear equation with a quadratic equation in two variables, solved graphically (and often checked with substitution). Intersection count can be 0, 1, or 2.

Quick example. Solve (y = x + 2) and (y = x^2 - 4): set (x + 2 = x^2 - 4) → (x^2 - x - 6 = 0) → ((x-3)(x+2)=0) → solutions ((3,5)) and ((-2,0)).

Graphically, enter both equations in the on-screen calculator and use intersect, checking that the window shows every crossing. For fuller worked examples, technology approximations with exponential partners, modeling constraints, and teaching scenarios about hidden intersections, see the dedicated section Systems with a Linear and a Quadratic Equation.