A collection of exercises on systems of equations using the methods of substitution, elimination, and row reduction. The set includes systems in two and three unknowns, special cases (dependent and inconsistent systems), and a few simple non-linear systems.
Before getting started, recall the main techniques:
• Substitution: solve one equation for a variable, then plug that expression into the other equation.
• Elimination: scale the equations so that the coefficients of one variable become equal or opposite, then add or subtract.
• Row reduction: bring the system to row-echelon form (especially handy for three-variable systems).
Always check your answers by substituting them back into the original equations.
Exercise 1 — level ★★☆☆☆
\[ \begin{cases} x + y = 5 \\ x - y = 1 \end{cases} \]
Answer
\( x = 3 \quad y = 2 \)
Solution
Elimination by addition
Adding the two equations term by term cancels \( y \):
\( (x + y) + (x - y) = 5 + 1 \implies 2x = 6 \implies x = 3 \)
Substituting into the first equation: \( 3 + y = 5 \implies y = 2 \).
Check
\( 3 + 2 = 5 \) and \( 3 - 2 = 1 \)
Answer: \(\boxed{x = 3 \quad y = 2}\)
Exercise 2 — level ★★☆☆☆
\[ \begin{cases} x - 2y = 0 \\ x + y = 6 \end{cases} \]
Answer
\( x = 4 \quad y = 2 \)
Solution
Substitution method
From the first equation, \( x = 2y \). Plugging into the second:
\( 2y + y = 6 \implies 3y = 6 \implies y = 2 \), and so \( x = 4 \).
Check
\( 4 - 4 = 0 \) and \( 4 + 2 = 6 \)
Answer: \(\boxed{x = 4 \quad y = 2}\)
Exercise 3 — level ★★☆☆☆
\[ \begin{cases} 3x + y = 10 \\ x + 3y = 6 \end{cases} \]
Answer
\( x = 3 \quad y = 1 \)
Solution
Substitution method
Solving the first equation for \( y \): \( y = 10 - 3x \). Plugging into the second:
\( x + 3(10 - 3x) = 6 \implies x + 30 - 9x = 6 \implies -8x = -24 \implies x = 3 \)
Then \( y = 10 - 9 = 1 \).
Check
\( 9 + 1 = 10 \) and \( 3 + 3 = 6 \)
Answer: \(\boxed{x = 3 \quad y = 1}\)
Exercise 4 — level ★★☆☆☆
\[ \begin{cases} 5x + 2y = 14 \\ x + y = 4 \end{cases} \]
Answer
\( x = 2 \quad y = 2 \)
Solution
Substitution method
From the second equation, \( x = 4 - y \). Plugging into the first:
\( 5(4 - y) + 2y = 14 \implies 20 - 5y + 2y = 14 \implies -3y = -6 \implies y = 2 \)
Therefore \( x = 4 - 2 = 2 \).
Check
\( 10 + 4 = 14 \) and \( 2 + 2 = 4 \)
Answer: \(\boxed{x = 2 \quad y = 2}\)
Exercise 5 — level ★★★☆☆
\[ \begin{cases} 2x - 3y = 1 \\ 4x + y = 9 \end{cases} \]
Answer
\( x = 2 \quad y = 1 \)
Solution
Elimination method
Multiply the second equation by 3 so that the coefficients of \( y \) are opposite:
\( \begin{cases} 2x - 3y = 1 \\ 12x + 3y = 27 \end{cases} \)
Adding: \( 14x = 28 \implies x = 2 \). It follows that \( y = 1 \).
Check
\( 4 - 3 = 1 \) and \( 8 + 1 = 9 \)
Answer: \(\boxed{x = 2 \quad y = 1}\)
Exercise 6 — level ★★★☆☆
\[ \begin{cases} 3x + 2y = 12 \\ 5x - 2y = 4 \end{cases} \]
Answer
\( x = 2 \quad y = 3 \)
Solution
Elimination method
The coefficients of \( y \) are already opposite, so we add the two equations:
\( 8x = 16 \implies x = 2 \). It follows that \( y = 3 \).
Check
\( 6 + 6 = 12 \) and \( 10 - 6 = 4 \)
Answer: \(\boxed{x = 2 \quad y = 3}\)
Exercise 7 — level ★★★☆☆
\[ \begin{cases} \dfrac{x}{3} + y = 3 \\ x + \dfrac{y}{2} = 4 \end{cases} \]
Answer
\( x = 3 \quad y = 2 \)
Solution
Clearing denominators
First equation ×3: \( x + 3y = 9 \)
Second equation ×2: \( 2x + y = 8 \)
From the first, \( x = 9 - 3y \). Plugging in: \( 2(9 - 3y) + y = 8 \implies y = 2 \), so \( x = 3 \).
Check
\( 1 + 2 = 3 \) and \( 3 + 1 = 4 \)
Answer: \(\boxed{x = 3 \quad y = 2}\)
Exercise 8 — level ★★★☆☆
\[ \begin{cases} 4x - 3y = -1 \\ 2x + 5y = 19 \end{cases} \]
Answer
\( x = 2 \quad y = 3 \)
Solution
Elimination method
Second equation ×2: \( 4x + 10y = 38 \). Subtracting the first:
\( 13y = 39 \implies y = 3 \). Hence \( x = 2 \).
Check
\( 8 - 9 = -1 \) and \( 4 + 15 = 19 \)
Answer: \(\boxed{x = 2 \quad y = 3}\)
Exercise 9 — level ★★★☆☆
\[ \begin{cases} 2x + 3y = 6 \\ 4x + 6y = 12 \end{cases} \]
Answer
Infinitely many solutions
Solution
Analysis of the system
Multiplying the first equation by 2 yields the second equation: the two are equivalent (they describe the same line).
The system is dependent. Solutions: \( x = \frac{6 - 3t}{2} \), \( y = t \) with \( t \in \mathbb{R} \).
Answer: \(\boxed{\text{Infinitely many solutions: } x = \dfrac{6-3t}{2},\ y = t \ (t \in \mathbb{R})}\)
Exercise 10 — level ★★★☆☆
\[ \begin{cases} 3x - y = 5 \\ 6x - 2y = 8 \end{cases} \]
Answer
No solution
Solution
Analysis of the system
Multiplying the first equation by 2 gives \( 6x - 2y = 10 \), which contradicts the second equation.
The two lines are parallel and distinct → inconsistent system.
Answer: \(\boxed{\text{Inconsistent system — no solution}}\)
Exercise 11 — level ★★★☆☆
\[ \begin{cases} \dfrac{x}{2} + \dfrac{y}{3} = 4 \\ \dfrac{x}{4} - \dfrac{y}{6} = 0 \end{cases} \]
Answer
\( x = 4 \quad y = 6 \)
Solution
Clearing denominators
First equation ×6: \( 3x + 2y = 24 \)
Second equation ×12: \( 3x - 2y = 0 \)
Elimination method
Adding the two equations cancels \( y \):
\( 6x = 24 \implies x = 4 \)
Substituting into the second: \( 12 - 2y = 0 \implies y = 6 \).
Check
\( \dfrac{4}{2} + \dfrac{6}{3} = 2 + 2 = 4 \) and \( \dfrac{4}{4} - \dfrac{6}{6} = 1 - 1 = 0 \)
Answer: \(\boxed{x = 4 \quad y = 6}\)
Exercise 12 — level ★★★☆☆
\[ \begin{cases} 7x - 2y = 16 \\ 3x + 4y = 2 \end{cases} \]
Answer
\( x = 2 \quad y = -1 \)
Solution
Elimination method
Multiply the first equation by 2 so that the coefficients of \( y \) become opposite:
\( \begin{cases} 14x - 4y = 32 \\ 3x + 4y = 2 \end{cases} \)
Adding: \( 17x = 34 \implies x = 2 \).
Substituting into the second equation: \( 6 + 4y = 2 \implies y = -1 \).
Check
\( 14 + 2 = 16 \) and \( 6 - 4 = 2 \)
Answer: \(\boxed{x = 2 \quad y = -1}\)
Exercise 13 — level ★★★☆☆
\[ \begin{cases} 4x + 3y = 17 \\ 5x - 2y = 4 \end{cases} \]
Answer
\( x = 2 \quad y = 3 \)
Solution
Elimination method (scaling both equations)
To eliminate \( y \), multiply the first equation by 2 and the second by 3:
\( \begin{cases} 8x + 6y = 34 \\ 15x - 6y = 12 \end{cases} \)
Adding: \( 23x = 46 \implies x = 2 \).
Substituting into the first: \( 8 + 3y = 17 \implies y = 3 \).
Check
\( 8 + 9 = 17 \) and \( 10 - 6 = 4 \)
Answer: \(\boxed{x = 2 \quad y = 3}\)
Exercise 14 — level ★★★☆☆
\[ \begin{cases} 2(x + 1) - y = 5 \\ x - (y - 3) = 4 \end{cases} \]
Answer
\( x = 2 \quad y = 1 \)
Solution
Preliminary simplification
Expanding the parentheses:
\( 2x + 2 - y = 5 \implies 2x - y = 3 \)
\( x - y + 3 = 4 \implies x - y = 1 \)
Elimination method
Subtracting the second from the first:
\( (2x - y) - (x - y) = 3 - 1 \implies x = 2 \)
Then \( y = x - 1 = 1 \).
Check
\( 2(3) - 1 = 5 \) and \( 2 - (1 - 3) = 4 \)
Answer: \(\boxed{x = 2 \quad y = 1}\)
Exercise 15 — level ★★★☆☆
\[ \begin{cases} -2x + 3y = -1 \\ 5x - y = 9 \end{cases} \]
Answer
\( x = 2 \quad y = 1 \)
Solution
Substitution method
From the second equation, \( y = 5x - 9 \). Plugging into the first:
\( -2x + 3(5x - 9) = -1 \implies -2x + 15x - 27 = -1 \implies 13x = 26 \implies x = 2 \)
Then \( y = 10 - 9 = 1 \).
Check
\( -4 + 3 = -1 \) and \( 10 - 1 = 9 \)
Answer: \(\boxed{x = 2 \quad y = 1}\)
Exercise 16 — level ★★★★☆
A father is 30 years older than his son. Ten years from now, the father's age will be twice his son's age. Find their current ages.
Answer
Father: 50 years old; son: 20 years old
Solution
Setting up the system
Let \( p \) be the father's age and \( s \) the son's age:
\[ \begin{cases} p = s + 30 \\ p + 10 = 2(s + 10) \end{cases} \]
Substitution method
Plugging \( p = s + 30 \) into the second equation:
\( (s + 30) + 10 = 2s + 20 \implies s + 40 = 2s + 20 \implies s = 20 \)
Hence \( p = 20 + 30 = 50 \).
Check
Current age difference: \( 50 - 20 = 30 \). Ten years on: \( 60 = 2 \cdot 30 \) ✓
Answer: \(\boxed{\text{Father: } 50 \text{ y/o; son: } 20 \text{ y/o}}\)
Exercise 17 — level ★★★★☆
\[ \begin{cases} 7x + 3y = 27 \\ 2x + 5y = 16 \end{cases} \]
Answer
\( x = 3 \quad y = 2 \)
Solution
Elimination method (scaling both equations)
To eliminate \( y \), multiply the first equation by 5 and the second by 3:
\( \begin{cases} 35x + 15y = 135 \\ 6x + 15y = 48 \end{cases} \)
Subtracting: \( 29x = 87 \implies x = 3 \).
Substituting back into the first original equation: \( 21 + 3y = 27 \implies y = 2 \).
Check
\( 21 + 6 = 27 \) and \( 6 + 10 = 16 \)
Answer: \(\boxed{x = 3 \quad y = 2}\)
Exercise 18 — level ★★★★☆
\[ \begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ x + y - z = 0 \end{cases} \]
Answer
\( x = 1 \quad y = 2 \quad z = 3 \)
Solution
Elimination method
Subtracting the second equation from the first:
\( (x + y + z) - (x - y + z) = 6 - 2 \implies 2y = 4 \implies y = 2 \)
Subtracting the third from the first:
\( (x + y + z) - (x + y - z) = 6 - 0 \implies 2z = 6 \implies z = 3 \)
Substituting into the first: \( x + 2 + 3 = 6 \implies x = 1 \).
Check
\( 1 + 2 + 3 = 6 \) ✓, \( 1 - 2 + 3 = 2 \) ✓, \( 1 + 2 - 3 = 0 \) ✓
Answer: \(\boxed{x = 1 \quad y = 2 \quad z = 3}\)
Exercise 19 — level ★★★★☆
\[ \begin{cases} x + 2y + z = 7 \\ 2x + y + z = 8 \\ x + y + 2z = 9 \end{cases} \]
Answer
\( x = 2 \quad y = 1 \quad z = 3 \)
Solution
Elimination method
Subtracting the first equation from the second:
\( x - y = 1 \quad (\text{A}) \)
Subtracting the first from the third:
\( -y + z = 2 \implies z = y + 2 \quad (\text{B}) \)
Substituting \( z = y + 2 \) and \( x = y + 1 \) (from A) into the first equation:
\( (y + 1) + 2y + (y + 2) = 7 \implies 4y + 3 = 7 \implies y = 1 \)
Therefore \( x = 2 \) and \( z = 3 \).
Check
\( 2 + 2 + 3 = 7 \) ✓, \( 4 + 1 + 3 = 8 \) ✓, \( 2 + 1 + 6 = 9 \) ✓
Answer: \(\boxed{x = 2 \quad y = 1 \quad z = 3}\)
Exercise 20 — level ★★★★☆
\[ \begin{cases} x + y + z = 6 \\ 2x + y - z = 1 \\ x - y + 2z = 5 \end{cases} \]
Answer
\( x = 1 \quad y = 2 \quad z = 3 \)
Solution
Elimination method
Subtracting the first equation from the second:
\( x - 2z = -5 \implies x = 2z - 5 \quad (\text{A}) \)
Adding the first and the third:
\( 2x + 3z = 11 \quad (\text{B}) \)
Plugging (A) into (B): \( 2(2z - 5) + 3z = 11 \implies 7z = 21 \implies z = 3 \).
Then \( x = 6 - 5 = 1 \), and from the first equation \( y = 6 - 1 - 3 = 2 \).
Check
\( 1 + 2 + 3 = 6 \) ✓, \( 2 + 2 - 3 = 1 \) ✓, \( 1 - 2 + 6 = 5 \) ✓
Answer: \(\boxed{x = 1 \quad y = 2 \quad z = 3}\)
Exercise 21 — level ★★★★☆
\[ \begin{cases} x + 2y + z = 8 \\ 2x + y + 3z = 9 \\ 3x + 4y + 2z = 20 \end{cases} \]
Answer
\( x = 4 \quad y = \dfrac{11}{5} \quad z = -\dfrac{2}{5} \)
Solution
Eliminating \( x \)
Form the combination \( 2 \cdot \text{Eq.}_1 - \text{Eq.}_2 \) to eliminate \( x \):
\( 2(x + 2y + z) - (2x + y + 3z) = 16 - 9 \implies 3y - z = 7 \quad (\text{A}) \)
Now form \( 3 \cdot \text{Eq.}_1 - \text{Eq.}_3 \):
\( 3(x + 2y + z) - (3x + 4y + 2z) = 24 - 20 \implies 2y + z = 4 \quad (\text{B}) \)
Solving the 2×2 system
Adding (A) and (B):
\( 5y = 11 \implies y = \dfrac{11}{5} \)
From (B): \( z = 4 - 2 \cdot \dfrac{11}{5} = \dfrac{20 - 22}{5} = -\dfrac{2}{5} \).
From the first original equation: \( x = 8 - 2y - z = 8 - \dfrac{22}{5} + \dfrac{2}{5} = 8 - 4 = 4 \).
Check
Eq. 1: \( 4 + \dfrac{22}{5} - \dfrac{2}{5} = 4 + 4 = 8 \) ✓
Eq. 2: \( 8 + \dfrac{11}{5} - \dfrac{6}{5} = 8 + 1 = 9 \) ✓
Eq. 3: \( 12 + \dfrac{44}{5} - \dfrac{4}{5} = 12 + 8 = 20 \) ✓
Answer: \(\boxed{x = 4 \quad y = \dfrac{11}{5} \quad z = -\dfrac{2}{5}}\)
Exercise 22 — level ★★★★★
\[ \begin{cases} 2x + y - z = 5 \\ x + 3y + 2z = 8 \\ 3x + 2y + 4z = 15 \end{cases} \]
Answer
\( x = \dfrac{67}{25} \quad y = \dfrac{23}{25} \quad z = \dfrac{32}{25} \)
Solution
Eliminating \( z \)
Form \( 2 \cdot \text{Eq.}_1 + \text{Eq.}_2 \) to eliminate \( z \):
\( 2(2x + y - z) + (x + 3y + 2z) = 10 + 8 \implies 5x + 5y = 18 \quad (\text{A}) \)
Now form \( 4 \cdot \text{Eq.}_1 + \text{Eq.}_3 \):
\( 4(2x + y - z) + (3x + 2y + 4z) = 20 + 15 \implies 11x + 6y = 35 \quad (\text{B}) \)
Solving the 2×2 system
From (A): \( x + y = \dfrac{18}{5} \), so \( y = \dfrac{18}{5} - x \).
Plugging into (B):
\( 11x + 6\left(\dfrac{18}{5} - x\right) = 35 \implies 11x + \dfrac{108}{5} - 6x = 35 \implies 5x = 35 - \dfrac{108}{5} = \dfrac{67}{5} \)
Hence \( x = \dfrac{67}{25} \).
Then \( y = \dfrac{18}{5} - \dfrac{67}{25} = \dfrac{90 - 67}{25} = \dfrac{23}{25} \).
From the first equation: \( z = 2x + y - 5 = \dfrac{134}{25} + \dfrac{23}{25} - \dfrac{125}{25} = \dfrac{32}{25} \).
Check
Eq. 2: \( \dfrac{67}{25} + \dfrac{69}{25} + \dfrac{64}{25} = \dfrac{200}{25} = 8 \) ✓
Eq. 3: \( \dfrac{201}{25} + \dfrac{46}{25} + \dfrac{128}{25} = \dfrac{375}{25} = 15 \) ✓
Answer: \(\boxed{x = \dfrac{67}{25} \quad y = \dfrac{23}{25} \quad z = \dfrac{32}{25}}\)
Exercise 23 — level ★★★★★
\[ \begin{cases} x + y = 6 \\ 2x + ky = 12 \end{cases} \]
Answer
Depends on the value of \( k \)
Solution
Discussion of the parameter
Substituting \( x = 6 - y \) yields \( (k - 2)y = 0 \).
- If \( k \neq 2 \): unique solution \( x = 6 \), \( y = 0 \) (consistent and independent system)
- If \( k = 2 \): infinitely many solutions \( (x = 6 - t,\ y = t) \) (consistent and dependent system)
Answer: \(\boxed{\text{Unique solution if } k \neq 2;\ \text{infinitely many solutions if } k=2}\)
Exercise 24 — level ★★★★★
\[ \begin{cases} x + y = 5 \\ x^2 + y^2 = 13 \end{cases} \]
Answer
\( (x,y) = (2,3) \) or \( (3,2) \)
Solution
Combined approach
From the first equation, \( y = 5 - x \). Plugging into the second: \( x^2 + (5 - x)^2 = 13 \implies x^2 - 5x + 6 = 0 \).
Solutions: \( x=2 \) (\( y=3 \)) and \( x=3 \) (\( y=2 \)).
Answer: \(\boxed{(x,y)=(2,3)\ \text{or}\ (3,2)}\)
Exercise 25 — level ★★★★★
\[ \begin{cases} x + y + 2z = 9 \\ 2x - y + z = 8 \\ x + 2y - z = 6 \end{cases} \]
Answer
\( x = \dfrac{49}{12} \quad y = \dfrac{7}{4} \quad z = \dfrac{19}{12} \)
Solution
Eliminating \( y \)
Adding \( \text{Eq.}_1 + \text{Eq.}_2 \) cancels \( y \):
\( (x + y + 2z) + (2x - y + z) = 9 + 8 \implies 3x + 3z = 17 \quad (\text{A}) \)
Now form \( 2 \cdot \text{Eq.}_2 + \text{Eq.}_3 \) to eliminate \( y \) once again:
\( 2(2x - y + z) + (x + 2y - z) = 16 + 6 \implies 5x + z = 22 \quad (\text{B}) \)
Solving the 2×2 system
From (A): \( z = \dfrac{17 - 3x}{3} \). Plugging into (B):
\( 5x + \dfrac{17 - 3x}{3} = 22 \implies 15x + 17 - 3x = 66 \implies 12x = 49 \implies x = \dfrac{49}{12} \)
Then \( z = \dfrac{17 - \frac{49}{4}}{3} = \dfrac{\frac{68 - 49}{4}}{3} = \dfrac{19}{12} \).
From the first original equation:
\( y = 9 - x - 2z = 9 - \dfrac{49}{12} - \dfrac{38}{12} = \dfrac{108 - 87}{12} = \dfrac{21}{12} = \dfrac{7}{4} \)
Check
Eq. 2: \( \dfrac{98}{12} - \dfrac{21}{12} + \dfrac{19}{12} = \dfrac{96}{12} = 8 \) ✓
Eq. 3: \( \dfrac{49}{12} + \dfrac{42}{12} - \dfrac{19}{12} = \dfrac{72}{12} = 6 \) ✓
Answer: \(\boxed{x = \dfrac{49}{12} \quad y = \dfrac{7}{4} \quad z = \dfrac{19}{12}}\)