4.3 Systems of Equations

Understanding Systems of Equations

A system of linear equations consists of two or more equations with the same set of variables. The solution to a system of equations in two variables is an ordered pair $(x, y)$ that satisfies both equations simultaneously. Graphically, the solution is the coordinates of the point where the lines intersect.

On the SHSAT, you will encounter three types of systems, categorized by the number of solutions they possess:

• One Solution: The lines have different slopes and intersect at exactly one point $(x, y)$. This is the most common scenario.
• No Solution: The lines are parallel. They have the exact same slope but different $y$-intercepts. Since parallel lines never intersect, there is no coordinate pair that satisfies both equations.
• Infinitely Many Solutions: The equations represent the exact same line. They have the same slope and same $y$-intercept. Any point on the line is a solution.

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Solving Systems by Substitution

The substitution method is best used when one of the equations can be easily solved for a single variable (i.e., the coefficient of the variable is $1$ or $-1$).

Step-by-Step Substitution Method

1. Isolate one variable: Solve one of the equations for either $x$ or $y$ in terms of the other variable.
2. Substitute: Replace that variable in the other equation with the expression you found. This creates an equation with only one variable.
3. Solve: Solve this new single-variable equation.
4. Back-Substitute: Plug the numerical value you found back into your original isolation expression to find the value of the second variable.
5. Verify: Check your solution by plugging $(x, y)$ back into both original equations.

• Example: Solve the system: $2x + y = 7$ $3x - 2y = 7$
  • Step 1: The variable $y$ in the first equation has a coefficient of $1$. Solve for $y$: $y = 7 - 2x$
  • Step 2: Substitute $7 - 2x$ for $y$ in the second equation: $3x - 2(7 - 2x) = 7$
  • Step 3: Solve the equation: $3x - 14 + 4x = 7$ $7x - 14 = 7$ $7x = 21$ $x = 3$
  • Step 4: Back-substitute $x = 3$ into the equation $y = 7 - 2x$: $y = 7 - 2(3) = 1$ The solution is the ordered pair $(3, 1)$.

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Solving Systems by Elimination

The elimination method (also called the addition method) is best used when equations are in standard form ($Ax + By = C$) and coefficients of one variable can easily be made opposites.

Step-by-Step Elimination Method

1. Align equations: Make sure both equations are written in standard form ($Ax + By = C$).
2. Create opposites: Multiply one or both equations by constants so that the coefficients of either $x$ or $y$ are opposites (e.g., $4y$ and $-4y$).
3. Add equations: Add the two equations together. The variable with opposite coefficients will cancel out (be eliminated), leaving a single-variable equation.
4. Solve: Solve the resulting equation.
5. Back-Substitute: Substitute that value back into either original equation to find the other variable.

• Example: Solve the system: $3x + 4y = 10$ $2x - 3y = 1$
  • Step 1: The equations are aligned.
  • Step 2: To eliminate $y$, find the LCM of the coefficients $4$ and $3$, which is $12$. Multiply the first equation by $3$ and the second equation by $4$: $3(3x + 4y = 10) \Rightarrow 9x + 12y = 30$ $4(2x - 3y = 1) \Rightarrow 8x - 12y = 4$
  • Step 3: Add the two equations: $(9x + 8x) + (12y - 12y) = 30 + 4$ $17x = 34$
  • Step 4: Solve for $x$: $x = 2$
  • Step 5: Substitute $x = 2$ back into the first equation: $3(2) + 4y = 10 \Rightarrow 6 + 4y = 10 \Rightarrow 4y = 4 \Rightarrow y = 1$ The solution is $(2, 1)$.

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SHSAT Application: System Word Problems

Many word problems on the SHSAT require you to set up and solve a system of linear equations. A classic example is the "Tickets" or "Coins" problem, where you must track two separate quantities: a total count and a total monetary value.

Solving a Ticket Word Problem

Problem: A high school theater sold a total of 150 tickets to a play. Student tickets sold for $5 each, and adult tickets sold for $10 each. If the total revenue from ticket sales was $1,100, how many adult tickets were sold?

1. Define the variables: Let $s$ be the number of student tickets sold. Let $a$ be the number of adult tickets sold.
2. Set up the system: The first equation represents the total number of tickets: $s + a = 150$ The second equation represents the total monetary value: $5s + 10a = 1100$
3. Solve the system: Use substitution. Solve the first equation for $s$: $s = 150 - a$ Substitute this expression into the value equation: $5(150 - a) + 10a = 1100$ $750 - 5a + 10a = 1100$ $750 + 5a = 1100$ $5a = 350$ $a = 70$ So, 70 adult tickets were sold. You can verify that $s = 150 - 70 = 80$ student tickets. The total value is $5(80) + 10(70) = 400 + 700 = 1100$, which matches.