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Systems of Linear Equations

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SATMath800 Algebra 04 — Systems of Linear Equations
Algebra · Area 04

Systems of Linear Equations

Two equations can describe the same two quantities. A system asks where those equations agree. Learn to solve systems by substitution, elimination, graphing, and strategic choice—and understand what the solution means.

Substitution Elimination Graphing One solution No solution Infinitely many solutions

1. What is a system of linear equations?

A system is a set of two or more equations that must be true at the same time. For a system of two linear equations in two variables, the solution is an ordered pair \((x,y)\) that makes both equations true.

\[\begin{aligned}x+y&=7\\2x-y&=5\end{aligned}\]

One equation gives a collection of possible points. The system asks for the point—or points—that belong to both equations.

SAT idea: A system is not two separate equations to solve independently. Look for the values of the variables that satisfy both equations together.

2. The coordinate-plane meaning of a system

Each linear equation can be graphed as a line. The solutions of the system are the points where the two graphs overlap.

x y (2, 1)
One intersection
One common point → one solution.
Parallel lines
No common point → no solution.
Same line
Every point is common → infinitely many solutions.

3. Solve by substitution

Substitution is especially useful when one equation already gives one variable in terms of the other.

\[y=2x+1\]

Suppose the system is:

\[\begin{aligned}y&=2x+1\\x+y&=10\end{aligned}\]
Substitute \(2x+1\) for \(y\):
\(x+(2x+1)=10\)
Combine like terms:
\(3x+1=10\)
Subtract 1 from both sides:
\(3x+1-1=10-1\)
\(3x=9\)
Divide both sides by 3:
\(\frac{3x}{3}=\frac{9}{3}\)
\(x=3\)
Substitute \(x=3\) into \(y=2x+1\):
\(y=2*(3)+1=7\)
Solution: \((x,y)=(3,7)\). Verify the result in both original equations before moving on.

4. Solve by elimination

Elimination is useful when adding or subtracting the equations can remove one variable.

\[\begin{aligned}2x+3y&=13\\2x-y&=5\end{aligned}\]
Subtract the second equation from the first so the \(2x\) terms cancel:
\((2x+3y)-(2x-y)=13-5\)
Simplify:
\(4y=8\)
Divide by 4:
\(\frac{4y}{4}=\frac{8}{4}\)
\(y=2\)
Substitute \(y=2\) into \(2x-y=5\):
\(2x-2=5\)
Add 2 to both sides:
\(2x-2+2=5+2\)
\(2x=7\)
Divide by 2:
\(\frac{2x}{2}=\frac{7}{2}\)
\(x=\frac{7}{2}\)
Solution: \((x,y)=\left(\frac{7}{2},2\right)\).

5. When the coefficients do not cancel immediately

Sometimes you first multiply one or both equations by a constant so that a variable has matching or opposite coefficients.

\[\begin{aligned}2x+3y&=12\\4x-y&=10\end{aligned}\]
Neither variable has opposite coefficients yet. Multiply the second equation by 3 so the \(y\)-terms become opposites.
\(3*(4x-y)=3*(10)\)
\(12x-3y=30\)
Add the first equation to the new equation so the \(y\)-terms cancel:
\((2x+3y)+(12x-3y)=12+30\)
\(14x=42\)
Divide by 14:
\(\frac{14x}{14}=\frac{42}{14}\)
\(x=3\)
Substitute \(x=3\) into \(4x-y=10\):
\(4*(3)-y=10\)
\(12-y=10\)
\(-y=-2\)
\(y=2\)

6. Graphing gives the solution visually

When two lines intersect at one point, the coordinates of that intersection are the solution to the system.

For example, if the graph shows an intersection at \((2,1)\), then the system has the solution \(x=2\), \(y=1\).

Important: The intersection is not just a visual feature. It represents the values that make both equations true.

7. One solution, no solution, or infinitely many?

Every system of two linear equations falls into one of three cases.

Exactly one solution
Different slopes → the lines intersect once.
No solution
Same slope, different intercepts → parallel lines.
Infinitely many solutions
The equations describe the same line.

Writing both equations in slope-intercept form can make the second and third cases easy to recognize.

A quick algorithm

When a system is given as equations rather than a graph, use this two-step test.

Step 1: Compare the slopes. Rewrite both equations as \(y=m*x+b\) and compare the values of \(m\).
If the slopes are different, the system has one solution. The lines must intersect once.
Step 2: If the slopes are the same, compare the y-intercepts.
Same slope and same intercept → the same line → infinitely many solutions.
Same slope and different intercepts → parallel lines → no solution.
\[m_1\ne m_2 \;\rightarrow\; \text{one solution}\] \[m_1=m_2,\ b_1=b_2 \;\rightarrow\; \text{infinitely many solutions}\] \[m_1=m_2,\ b_1\ne b_2 \;\rightarrow\; \text{no solution}\]
SAT shortcut: Different slopes → one solution. Same slope → compare the intercepts.

8. No solution: parallel lines

x y line 1 line 2
\[y=2x+1\qquad y=2x-4\]

The slopes are equal, but the y-intercepts are different. The lines never meet, so there is no ordered pair that satisfies both equations.

9. Infinitely many solutions: the same line

x y same line
\[2x+4y=10\qquad x+2y=5\]

The first equation is exactly 2 times the second equation. They represent the same line, so every point on that line satisfies both equations.

SAT shortcut: If multiplying one entire equation by a nonzero constant produces the other equation, the system has infinitely many solutions.

10. Systems in context

A context often gives two relationships involving the same quantities. Translate each relationship into an equation, then solve the system.

Example: A school sold 42 tickets. Student tickets cost $6 each and adult tickets cost $10 each. The total revenue was $324. Let \(s\) be student tickets and \(a\) be adult tickets.
\[\begin{aligned}s+a&=42\\6s+10a&=324\end{aligned}\]
From the first equation, write \(s=42-a\).
Substitute into the revenue equation:
\(6*(42-a)+10a=324\).
Expand and combine:
\(252-6a+10a=324\), so \(4a=72\).
Solve: \(a=18\). Then \(s=42-18=24\).
Check the context: \(24+18=42\) tickets and \(6*(24)+10*(18)=324\).

11. A reliable SAT systems strategy

Observe: Look at the coefficients and forms of the equations.

Identify: Decide whether substitution, elimination, or graphing is most efficient.

Decompose: Isolate one variable or create matching coefficients.

Reason: Solve for one variable, then use it to find the other.

Solve: Write the ordered pair or the requested quantity.

Verify: Substitute the result into both original equations or check it against the context.

12. Original SAT-style practice

Question 1 · Solve by substitution

What is the solution \((x,y)\) to the system?

\[\begin{aligned}y&=x+4\\2x+y&=13\end{aligned}\]
  • A. \((2,6)\)
  • B. \((3,7)\)
  • C. \((4,8)\)
  • D. \((5,9)\)
Answer: B.
Substitute \(x+4\) for \(y\):
\(2x+(x+4)=13\)
Combine like terms:
\(3x+4=13\)
Subtract 4 from both sides:
\(3x+4-4=13-4\)
\(3x=9\)
Divide both sides by 3:
\(\frac{3x}{3}=\frac{9}{3}\)
\(x=3\)
Substitute \(x=3\) into the first equation:
\(y=3+4=7\)
Write the solution as an ordered pair:
\((x,y)=(3,7)\)
Verify: \(3+4=7\), and \(2*(3)+7=13\). Both original equations are satisfied.
Question 2 · Solve by elimination

What is the value of \(x\) in the system?

\[\begin{aligned}3x+2y&=17\\5x-2y&=23\end{aligned}\]
  • A. 3
  • B. 4
  • C. 5
  • D. 6
Answer: C.
Add the equations so the \(y\)-terms cancel: \(8x=40\) .
Divide by 8: \(\frac{8x}{8}=\frac{40}{8}\) , so \(x=5\).
Question 3 · Identify the number of solutions

How many solutions does the system have?

\[\begin{aligned}2x+4y&=10\\x+2y&=7\end{aligned}\]
  • A. Zero
  • B. Exactly one
  • C. Exactly two
  • D. Infinitely many
Answer: A.
Multiply the second equation by 2: \(2x+4y=14\).
The first equation says \(2x+4y=10\).
The same expression cannot equal both 10 and 14. Therefore the system has no solution .
Question 4 · Identify infinitely many solutions

How many solutions does the system have?

\[\begin{aligned}4x-6y&=18\\2x-3y&=9\end{aligned}\]
  • A. Zero
  • B. Exactly one
  • C. Exactly two
  • D. Infinitely many
Answer: D.
Multiply the second equation by 2: \(4x-6y=18\).
This is exactly the first equation, so both equations describe the same line.
Every point on that line satisfies both equations. There are infinitely many solutions.
Question 5 · Choose an efficient method

Which value of \(y\) satisfies the system?

\[\begin{aligned}x&=3y-2\\4x+y&=18\end{aligned}\]
  • A. 1
  • B. 2
  • C. 3
  • D. 4
Answer: B.
The first equation already gives \(x\), so substitution is efficient.
Substitute: \(4*(3y-2)+y=18\).
Expand: \(12y-8+y=18\).
Combine: \(13y=26\).
Thus \(y=\frac{26}{13}=2\).
Question 6 · Find the intersection from a graph

The graph below represents a system of two linear equations. What is the solution to the system?

x y (2, 1)
  • A. \((1,2)\)
  • B. \((2,1)\)
  • C. \((2,2)\)
  • D. \((3,1)\)
Answer: B.

The solution is the point where the two lines intersect: \((2,1)\).

Question 7 · No solution from slopes

Which statement about the system is true?

\[\begin{aligned}y&=\frac{3}{4}x+2\\y&=\frac{3}{4}x-5\end{aligned}\]
  • A. It has exactly one solution.
  • B. It has no solution.
  • C. It has exactly two solutions.
  • D. It has infinitely many solutions.
Answer: B.
Both lines have slope \(\frac{3}{4}\).
Their y-intercepts are different: 2 and −5.
Therefore the lines are parallel and never intersect.
Question 8 · A system in context

A theater sold 80 tickets. Adult tickets cost $12 and student tickets cost $7. The theater collected $785. How many student tickets were sold?

\[\begin{aligned}a+s&=80\\12a+7s&=785\end{aligned}\]
  • A. 35
  • B. 40
  • C. 45
  • D. 50
Answer: A.
From \(a+s=80\), write \(a=80-s\).
Substitute: \(12*(80-s)+7s=785\).
Expand: \(960-12s+7s=785\).
Combine: \(960-5s=785\).
Subtract 960: \(-5s=-175\).
Divide by −5: \(s=35\).
Question 9 · Recognize equivalent equations

For which value of \(k\) does the system have infinitely many solutions?

\[\begin{aligned}3x-2y&=7\\-6x+4y&=k\end{aligned}\]
  • A. −14
  • B. −7
  • C. 7
  • D. 14
Answer: A.
Multiply the first equation by −2: \(-6x+4y=-14\).
For the equations to be equivalent, the second equation must have the same constant.
Therefore \(k=-14\).
Question 10 · Verify a proposed solution

A student claims that \((2,3)\) is the solution to the system. Is the claim correct?

\[\begin{aligned}x+2y&=8\\3x-y&=3\end{aligned}\]
  • A. Yes, because the point satisfies both equations.
  • B. No, because it satisfies neither equation.
  • C. No, because it satisfies only the first equation.
  • D. No, because it satisfies only the second equation.
Answer: A.
First equation: \(2+2*(3)=8\), true.
Second equation: \(3*(2)-3=3\), true.
Because both equations are true, \((2,3)\) is the solution.
Question 11 · Read the solution from a graph

The graph below represents a system of two linear equations. What is the solution to the system?

(4, 3) x y
  • A. \((3,4)\)
  • B. \((4,3)\)
  • C. \((4,2)\)
  • D. \((5,3)\)
Answer: B.
A solution to a system must satisfy both equations.
On the graph, the two lines meet at the marked point \((4,3)\).
Therefore, the solution is \((4,3)\).
Question 12 · Determine the number of solutions from a graph

The graph below shows the two lines in a system. How many solutions does the system have?

Line 1 Line 2 x y
  • A. 0
  • B. 1
  • C. 2
  • D. Infinitely many
Answer: A.
The two lines have the same slope, so they are parallel.
They are different lines, so they never intersect.
Therefore, the system has 0 solutions.
Question 13 · Recognize infinitely many solutions from a graph

The two equations in a system are shown on the graph. How many solutions does the system have?

y = −2x + 4 2y = −4x + 8 x y
  • A. 0
  • B. 1
  • C. 2
  • D. Infinitely many
Answer: D.
The two equations represent the same line.
Every point on that line satisfies both equations.
Therefore, the system has infinitely many solutions.
Question 14 · Find a coordinate from the intersection

The graph represents the system \(y=2x-1\) and \(x=3\). What is the value of \(y\) at the solution to the system?

(3, 5) 3 5 x y
  • A. 3
  • B. 4
  • C. 5
  • D. 6
Answer: C.
The vertical line gives \(x=3\).
Substitute \(x=3\) into \(y=2x-1\).
\(y=2*(3)-1=6-1=5\).
Therefore, the value of \(y\) is 5.
Question 15 · Interpret the intersection in context

Two internet plans are represented by the lines in the graph. The horizontal coordinate represents the number of months, and the vertical coordinate represents the total cost in dollars. At what total cost are the two plans equal?

(5, 50) Plan APlan B monthscost ($)
  • A. $40
  • B. $50
  • C. $60
  • D. $70
Answer: B.
The two plans have the same total cost at their point of intersection.
The graph marks the intersection at \((5,50)\).
The vertical coordinate is the total cost, so the two plans are equal at $50.

13. Mastery check

Substitution
Best when one variable is already isolated or easy to isolate.
Elimination
Best when coefficients can cancel with little work.
Graphing
Best when the intersection or relationship between the lines is the main idea.

Final SAT habit: Before doing algebra, inspect the structure of the system. The fastest correct method is often visible before you calculate.

SATMath800 · Algebra Learning System · Learn → Practice → Master → Perform

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