💡 8th Grade Math (Algebra I): Systems of Linear Equations Practice Questions

💡 Understanding the Substitution Method: This method involves solving one equation for one variable and then substituting that expression into the other equation.

• Step 1: Identify an isolated variable.
  The first equation, \(y = 3x - 5\), already has \(y\) isolated. This makes substitution straightforward!
• Step 2: Substitute the expression into the second equation.
  Substitute \((3x - 5)\) for \(y\) in the second equation: \[ 2x + (3x - 5) = 10 \]
• Step 3: Solve the resulting equation for \(x\).
  Combine like terms: \[ 5x - 5 = 10 \] Add 5 to both sides: \[ 5x = 15 \] Divide by 5: \[ x = 3 \]
• Step 4: Substitute the value of \(x\) back into one of the original equations to find \(y\).
  Using the first equation \(y = 3x - 5\): \[ y = 3(3) - 5 \] \[ y = 9 - 5 \] \[ y = 4 \]
• Step 5: Write the solution as an ordered pair.
  The solution is \((3, 4)\).

✅ Check your answer: Substitute \(x=3\) and \(y=4\) into both original equations.

• Equation 1: \(4 = 3(3) - 5 \implies 4 = 9 - 5 \implies 4 = 4\) (True)
• Equation 2: \(2(3) + 4 = 10 \implies 6 + 4 = 10 \implies 10 = 10\) (True)

Both equations hold true, so our solution is correct! 🎉

💡 Understanding the Elimination Method: This method involves adding or subtracting the equations to eliminate one of the variables.

• Step 1: Identify variables with opposite or matching coefficients.
  In this system, the \(y\) terms have coefficients of \(+1\) and \(-1\). If we add the equations, the \(y\) terms will cancel out!
• Step 2: Add the two equations together.
  \[ \begin{align*} (x + y) + (x - y) &= 7 + 3 \\ x + y + x - y &= 10 \\ 2x &= 10 \end{align*} \]
• Step 3: Solve the resulting equation for the remaining variable.
  \[ 2x = 10 \] Divide by 2: \[ x = 5 \]
• Step 4: Substitute the value of \(x\) back into one of the original equations to find \(y\).
  Using the first equation \(x + y = 7\): \[ 5 + y = 7 \] Subtract 5 from both sides: \[ y = 2 \]
• Step 5: Write the solution as an ordered pair.
  The solution is \((5, 2)\).

✅ Check your answer: Substitute \(x=5\) and \(y=2\) into both original equations.

• Equation 1: \(5 + 2 = 7 \implies 7 = 7\) (True)
• Equation 2: \(5 - 2 = 3 \implies 3 = 3\) (True)

Both equations hold true. Great job! 👍

💡 Strategy for Elimination: Sometimes you need to multiply one or both equations by a constant to create matching or opposite coefficients.

• Step 1: Choose a variable to eliminate.
  Let's choose to eliminate \(y\). The coefficients are \(+2\) and \(-4\). If we multiply the first equation by 2, the \(y\) coefficient will become \(+4\), which is opposite to \(-4\).
• Step 2: Multiply the first equation by 2.
  \[ 2(3x + 2y) = 2(10) \] \[ 6x + 4y = 20 \] The system now looks like this: \[ 6x + 4y = 20 \quad \text{(New Equation 1)} \] \[ x - 4y = -6 \quad \text{(Equation 2)} \]
• Step 3: Add the new Equation 1 and Equation 2.
  \[ \begin{align*} (6x + 4y) + (x - 4y) &= 20 + (-6) \\ 6x + 4y + x - 4y &= 14 \\ 7x &= 14 \end{align*} \]
• Step 4: Solve for \(x\).
  \[ 7x = 14 \] Divide by 7: \[ x = 2 \]
• Step 5: Substitute the value of \(x\) into one of the original equations to find \(y\).
  Using the second original equation \(x - 4y = -6\) because it's simpler: \[ 2 - 4y = -6 \] Subtract 2 from both sides: \[ -4y = -8 \] Divide by -4: \[ y = 2 \]
• Step 6: Write the solution as an ordered pair.
  The solution is \((2, 2)\).

✅ Self-check:

• Equation 1: \(3(2) + 2(2) = 6 + 4 = 10\) (True)
• Equation 2: \(2 - 4(2) = 2 - 8 = -6\) (True)

💡 Choosing the Right Variable to Isolate: Look for a variable with a coefficient of 1 or -1, as it simplifies the isolation step.

• Step 1: Isolate one variable in one of the equations.
  From the second equation, \(x - 2y = -4\), we can easily isolate \(x\) by adding \(2y\) to both sides: \[ x = 2y - 4 \]
• Step 2: Substitute this expression into the other equation.
  Substitute \((2y - 4)\) for \(x\) in the first equation \(2x + 3y = 13\): \[ 2(2y - 4) + 3y = 13 \]
• Step 3: Solve the resulting equation for \(y\).
  Distribute the 2: \[ 4y - 8 + 3y = 13 \] Combine like terms: \[ 7y - 8 = 13 \] Add 8 to both sides: \[ 7y = 21 \] Divide by 7: \[ y = 3 \]
• Step 4: Substitute the value of \(y\) back into the isolated expression for \(x\).
  Using \(x = 2y - 4\): \[ x = 2(3) - 4 \] \[ x = 6 - 4 \] \[ x = 2 \]
• Step 5: Write the solution as an ordered pair.
  The solution is \((2, 3)\).

✅ Verification:

• Equation 1: \(2(2) + 3(3) = 4 + 9 = 13\) (True)
• Equation 2: \(2 - 2(3) = 2 - 6 = -4\) (True)

💡 Key Concept: Number of Solutions
The number of solutions to a system of two linear equations depends on the relationship between their lines when graphed:

• One Solution: The lines intersect at exactly one point. (Different slopes)
• No Solution: The lines are parallel and never intersect. (Same slope, different y-intercepts)
• Infinitely Many Solutions: The lines are the same (coincident). (Same slope, same y-intercept)

• Step 1: Convert both equations to slope-intercept form (\(y = mx + b\)).
  Equation 1 is already in slope-intercept form: \[ y = 2x + 1 \] Here, the slope \(m_1 = 2\) and the y-intercept \(b_1 = 1\).
  Now, convert Equation 2: \(4x - 2y = -2\) Subtract \(4x\) from both sides: \[ -2y = -4x - 2 \] Divide all terms by -2: \[ y = \frac{-4x}{-2} + \frac{-2}{-2} \] \[ y = 2x + 1 \] Here, the slope \(m_2 = 2\) and the y-intercept \(b_2 = 1\).
• Step 2: Compare the slopes and y-intercepts.
  We found that \(m_1 = 2\) and \(m_2 = 2\). The slopes are the same.
  We also found that \(b_1 = 1\) and \(b_2 = 1\). The y-intercepts are the same.
• Step 3: Determine the number of solutions based on the comparison.
  Since both the slopes and the y-intercepts are identical, the two equations represent the same line. Therefore, there are infinitely many solutions. Every point on the line is a solution to the system.

📌 Reasoning: The two equations are equivalent, meaning they graph the exact same line. When two lines are coincident, they share every single point, leading to an infinite number of solutions. 🎉

💡 Translating Word Problems into Systems: Identify the unknown quantities and define variables for them. Then, find two different relationships (equations) between these variables.

• Step 1: Define the variables.
  Let \(b\) represent the number of bracelets Sarah sold.
  Let \(n\) represent the number of necklaces Sarah sold.
• Step 2: Formulate the equations.
  From the problem, we know two things:
  1. Total number of items: She sold a total of 20 items. \[ b + n = 20 \quad \text{(Equation 1)} \]
  2. Total money made: Bracelets cost 5, necklaces cost 8, and she made 130. \[ 5b + 8n = 130 \quad \text{(Equation 2)} \]
  Now we have a system of linear equations: \[ b + n = 20 \] \[ 5b + 8n = 130 \]
• Step 3: Solve the system (using substitution or elimination).
  Let's use the substitution method. From Equation 1, isolate \(b\): \[ b = 20 - n \] Substitute this into Equation 2: \[ 5(20 - n) + 8n = 130 \] Distribute: \[ 100 - 5n + 8n = 130 \] Combine like terms: \[ 100 + 3n = 130 \] Subtract 100 from both sides: \[ 3n = 30 \] Divide by 3: \[ n = 10 \] Now substitute \(n = 10\) back into \(b = 20 - n\): \[ b = 20 - 10 \] \[ b = 10 \]
• Step 4: State the answer in the context of the problem.
  Sarah sold 10 bracelets and 10 necklaces.

✅ Check:

• Total items: \(10 + 10 = 20\) (Correct)
• Total money: \(5(10) + 8(10) = 50 + 80 = 130\) (Correct)

💡 Understanding Line Relationships: The relationship between two lines (intersecting, parallel, or coincident) is determined by comparing their slopes and y-intercepts.

• Step 1: Determine the slope and y-intercept of Line A.
  Line A is given by \(y = -3x + 5\). Its slope \(m_A = -3\). Its y-intercept \(b_A = 5\).
• Step 2: Determine the slope of Line B.
  Line B passes through points \((x_1, y_1) = (1, 2)\) and \((x_2, y_2) = (3, -4)\). The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). \[ m_B = \frac{-4 - 2}{3 - 1} = \frac{-6}{2} = -3 \]
• Step 3: Compare the slopes of Line A and Line B.
  We have \(m_A = -3\) and \(m_B = -3\). Since the slopes are the same, the lines are either parallel or they are the same line.
• Step 4: Determine the y-intercept of Line B (if necessary).
  Since the slopes are the same, we need to check the y-intercepts. We can use the point-slope form \(y - y_1 = m(x - x_1)\) with one of the points from Line B, e.g., \((1, 2)\), and its slope \(m_B = -3\): \[ y - 2 = -3(x - 1) \] Distribute: \[ y - 2 = -3x + 3 \] Add 2 to both sides: \[ y = -3x + 5 \] So, the equation for Line B is \(y = -3x + 5\). Its y-intercept \(b_B = 5\).
• Step 5: Compare the y-intercepts and conclude.
  We have \(b_A = 5\) and \(b_B = 5\). Since both the slopes and the y-intercepts are identical, the two lines are the same line.

📌 Conclusion: Lines A and B are the same line. This means they have infinitely many solutions, as every point on one line is also on the other line. 🎉

💡 Setting up Equations from Real-World Scenarios: Define your variables clearly and write two equations based on the given totals (e.g., total quantity and total value).

• Step 1: Define the variables.
  Let \(A\) represent the number of adult tickets sold.
  Let \(C\) represent the number of child tickets sold.
• Step 2: Formulate the equations.
  From the problem:
  1. Total number of tickets: 300 tickets were sold. \[ A + C = 300 \quad \text{(Equation 1)} \]
  2. Total revenue: Adult tickets are 15, child tickets are 10, total revenue 4000. \[ 15A + 10C = 4000 \quad \text{(Equation 2)} \]
  Our system is: \[ A + C = 300 \] \[ 15A + 10C = 4000 \]
• Step 3: Solve the system (using elimination).
  Let's use elimination. To eliminate \(C\), multiply Equation 1 by \(-10\): \[ -10(A + C) = -10(300) \] \[ -10A - 10C = -3000 \quad \text{(New Equation 1)} \] Now, add New Equation 1 and Equation 2: \[ \begin{align*} (-10A - 10C) + (15A + 10C) &= -3000 + 4000 \\ -10A + 15A - 10C + 10C &= 1000 \\ 5A &= 1000 \end{align*} \] Divide by 5: \[ A = 200 \] Now substitute \(A = 200\) back into original Equation 1 \(A + C = 300\): \[ 200 + C = 300 \] Subtract 200 from both sides: \[ C = 100 \]
• Step 4: State the answer in the context of the problem.
  The theater sold 200 adult tickets and 100 child tickets.

✅ Verification:

• Total tickets: \(200 + 100 = 300\) (Correct)
• Total revenue: \(15(200) + 10(100) = 3000 + 1000 = 4000\) (Correct)