Digital SAT Maths -Unit 1 - 1.4 Systems of Linear Equations- Study Notes- New Syllabus
Digital SAT Maths -Unit 1 – 1.4 Systems of Linear Equations- Study Notes- New syllabus
Digital SAT Maths -Unit 1 – 1.4 Systems of Linear Equations- Study Notes – per latest Syllabus.
Key Concepts:
Graphical, substitution, elimination
One / no / infinite solutions
Word problems
Graphical, Substitution, and Elimination Methods
A system of linear equations consists of two equations involving the same variables. The solution is the ordered pair \( (x, y) \) that satisfies both equations simultaneously.
On the DIGITAL SAT, systems are very often written in standard form.
Standard Form
\( a_1x + b_1y = c_1 \)
\( a_2x + b_2y = c_2 \)
Each equation represents a straight line. The solution is the point where the two lines intersect.
Method 1: Graphical Method
Each equation represents a straight line. The solution is the point where the lines cross.
Steps:
- Graph both equations
- Find the intersection point
- That point is the solution
On the DIGITAL SAT, you are usually given the graph rather than asked to draw it.
Method 2: Substitution Method
Use substitution when one equation already isolates a variable.
Steps:
- Solve one equation for a variable
- Substitute into the other equation
- Solve for one variable
- Substitute back
Method 3: Elimination Method
Use elimination when coefficients can cancel.
Steps:
- Multiply equations if needed
- Add or subtract equations
- Solve
- Back-substitute
DIGITAL SAT Insight
The SAT often asks for only one value, such as \( x \), \( y \), or \( x + y \), instead of the ordered pair.
Example 1 (Graphical Method):
Solve the system graphically:
\( y = x + 1 \)
\( y = -x + 5 \)
▶️ Answer/Explanation
Step 1: Create a table for \( y = x + 1 \)
| \( x \) | \( y = x + 1 \) |
|---|---|
| 0 | 1 |
| 2 | 3 |
| 4 | 5 |
Step 2: Create a table for \( y = -x + 5 \)
| \( x \) | \( y = -x + 5 \) |
|---|---|
| 0 | 5 |
| 2 | 3 |
| 4 | 1 |
Plot the points from both tables and draw straight lines through them.
Both lines intersect at \( (2, 3) \).
Conclusion: The solution to the system is \( (2, 3) \).
Example 2 (Substitution Method):
Solve the system:
\( y = 3x + 2 \)
\( 2x + y = 14 \)
▶️ Answer/Explanation
Substitute \( y = 3x + 2 \):
\( 2x + (3x + 2) = 14 \)
\( 5x + 2 = 14 \)
\( 5x = 12 \)
\( x = \dfrac{12}{5} \)
Find \( y \):
\( y = 3\left(\dfrac{12}{5}\right) + 2 = \dfrac{36}{5} + \dfrac{10}{5} = \dfrac{46}{5} \)
Conclusion: The solution is \( \left(\dfrac{12}{5}, \dfrac{46}{5}\right) \).
Example 3 (Elimination Method):
Solve the system:
\( 3x + 2y = 16 \)
\( 3x – 2y = 4 \)
▶️ Answer/Explanation
Add equations:
\( 6x = 20 \)
\( x = \dfrac{10}{3} \)
Substitute into first equation:
\( 3\left(\dfrac{10}{3}\right) + 2y = 16 \)
\( 10 + 2y = 16 \)
\( 2y = 6 \)
\( y = 3 \)
Conclusion: The solution is \( \left(\dfrac{10}{3}, 3\right) \).
One, No, or Infinitely Many Solutions (Using Standard Form)
On the DIGITAL SAT, systems are often written in standard form:
\( a_1x + b_1y = c_1 \)
\( a_2x + b_2y = c_2 \)
You are frequently asked how many solutions the system has without fully solving it. This depends on the relationship between the coefficients.
Key Coefficient Rule
- If \( \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} \) → one solution
- If \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} \) → no solution
- If \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \) → infinitely many solutions
Why this works:
- Different ratios → different slopes → lines intersect
- Same ratios but different constants → parallel lines
- All ratios equal → same line
What Happens Algebraically
- One solution → you get a number for \( x \) and \( y \)
- No solution → false statement (example \( 4 = 9 \))
- Infinite solutions → true statement (example \( 7 = 7 \))
Example 1 (One Solution):
\( 2x + 3y = 11 \)
\( 4x – y = 5 \)
Determine the number of solutions.
▶️ Answer/Explanation
Compare ratios:
\( \dfrac{2}{4} = \dfrac{1}{2} \)
\( \dfrac{3}{-1} = -3 \)
Ratios are different.
Conclusion: The system has exactly one solution.
Example 2 (No Solution):
\( 3x + 6y = 9 \)
\( x + 2y = 1 \)
Determine the number of solutions.
▶️ Answer/Explanation
Ratios:
\( \dfrac{3}{1} = 3 \)
\( \dfrac{6}{2} = 3 \)
\( \dfrac{9}{1} = 9 \)
First two ratios equal but constant ratio different.
Conclusion: No solution (parallel lines).
Example 3 (Infinite Solutions):
\( 2x + 4y = 8 \)
\( x + 2y = 4 \)
Determine the number of solutions.
▶️ Answer/Explanation
Ratios:
\( \dfrac{2}{1} = 2 \)
\( \dfrac{4}{2} = 2 \)
\( \dfrac{8}{4} = 2 \)
All ratios equal.
Conclusion: Infinitely many solutions.
Word Problems
On the DIGITAL SAT, systems of equations most commonly appear as real-world modeling problems. You must translate the situation into two equations and then solve.
These problems usually involve two unknown quantities such as:
- two prices
- two speeds
- two plans
- two mixtures
How to Build the System
- Define variables clearly
- Use totals (total cost, total distance, total quantity)
- Write two independent equations
- Solve using substitution or elimination
- Interpret the answer in context
Common SAT Translation Patterns
- “total items” → first equation
- “total cost/value” → second equation
- “same time” or “meet” → distances equal
Example 1 (Tickets):
At a concert, adult tickets cost \$15 and student tickets cost \$9. A group purchased 20 tickets for \$228. How many student tickets were bought?
▶️ Answer/Explanation
Let
\( a = \) adult tickets
\( s = \) student tickets
Total tickets:
\( a + s = 20 \)
Total cost:
\( 15a + 9s = 228 \)
From first equation:
\( a = 20 – s \)
Substitute:
\( 15(20 – s) + 9s = 228 \)
\( 300 – 15s + 9s = 228 \)
\( 300 – 6s = 228 \)
\( -6s = -72 \)
\( s = 12 \)
Conclusion: 12 student tickets were purchased.
Example 2 (Two Speeds Meeting):
Two cars start 210 miles apart and drive toward each other. One travels 60 mph and the other 45 mph. After how many hours will they meet?
▶️ Answer/Explanation
Let \( t \) be hours traveled.
Distances:
Car 1: \( 60t \)
Car 2: \( 45t \)
They meet when total distance equals 210:
\( 60t + 45t = 210 \)
\( 105t = 210 \)
\( t = 2 \)
Conclusion: They meet after 2 hours.
Example 3 (Mixture):
A store mixes peanuts costing \$4 per kg with almonds costing \$10 per kg to make 12 kg of a \$6 per kg mixture. How many kilograms of peanuts are used?
▶️ Answer/Explanation
Let
\( p = \) kg peanuts
\( a = \) kg almonds
Total weight:
\( p + a = 12 \)
Total value:
\( 4p + 10a = 6(12) \)
\( 4p + 10a = 72 \)
From first equation:
\( a = 12 – p \)
Substitute:
\( 4p + 10(12 – p) = 72 \)
\( 4p + 120 – 10p = 72 \)
\( -6p = -48 \)
\( p = 8 \)
Conclusion: 8 kg of peanuts were used.