Collection of exercises on systems of equations using substitution, elimination and reduction methods. Includes systems with 2 and 3 variables, special cases (indeterminate and inconsistent systems) and some simple nonlinear systems.
Before starting, remember the main methods:
• Substitution: solve one equation for one variable and substitute it into the other equation.
• Elimination: multiply the equations to make the coefficients of one variable equal or opposite, then add or subtract them.
• Reduction: transform the system into row echelon form (especially useful with three variables).
Always verify the solutions 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 method (addition)
Adding the two equations member by member eliminates \( y \):
\( (x + y) + (x - y) = 5 + 1 \implies 2x = 6 \implies x = 3 \)
Substituting into the first equation: \( 3 + y = 5 \implies y = 2 \).
Verification
\( 3 + 2 = 5 \) and \( 3 - 2 = 1 \)
Final 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: \( x = 2y \). Substituting into the second:
\( 2y + y = 6 \implies 3y = 6 \implies y = 2 \), therefore \( x = 4 \).
Verification
\( 4 - 4 = 0 \) and \( 4 + 2 = 6 \)
Final 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
From the first: \( y = 10 - 3x \). Substituting into the second:
\( x + 3(10 - 3x) = 6 \implies x + 30 - 9x = 6 \implies -8x = -24 \implies x = 3 \)
Then \( y = 10 - 9 = 1 \).
Verification
\( 9 + 1 = 10 \) and \( 3 + 3 = 6 \)
Final 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: \( x = 4 - y \). Substituting into the first:
\( 5(4 - y) + 2y = 14 \implies 20 - 5y + 2y = 14 \implies -3y = -6 \implies y = 2 \)
Then \( x = 4 - 2 = 2 \).
Verification
\( 10 + 4 = 14 \) and \( 2 + 2 = 4 \)
Final 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
We multiply the second equation by 3 to make the coefficients of \( y \) opposite:
\( \begin{cases} 2x - 3y = 1 \\ 12x + 3y = 27 \end{cases} \)
Adding them: \( 14x = 28 \implies x = 2 \). Then \( y = 1 \).
Verification
\( 4 - 3 = 1 \) and \( 8 + 1 = 9 \)
Final 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. Adding the equations:
\( 8x = 16 \implies x = 2 \). Then \( y = 3 \).
Verification
\( 6 + 6 = 12 \) and \( 10 - 6 = 4 \)
Final 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 fractions
First equation ×3: \( x + 3y = 9 \)
Second equation ×2: \( 2x + y = 8 \)
From the first: \( x = 9 - 3y \). Substituting: \( 2(9 - 3y) + y = 8 \implies y = 2 \), therefore \( x = 3 \).
Verification
\( 1 + 2 = 3 \) and \( 3 + 1 = 4 \)
Final 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:
\( 13y = 39 \implies y = 3 \). Then \( x = 2 \).
Verification
\( 8 - 9 = -1 \) and \( 4 + 15 = 19 \)
Final 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
System analysis
Multiplying the first equation by 2 gives the second: the equations are equivalent (same line).
The system is indeterminate. Solutions: \( x = \frac{6 - 3t}{2} \), \( y = t \) with \( t \in \mathbb{R} \).
Final 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
System analysis
Multiplying the first by 2: \( 6x - 2y = 10 \), which contradicts the second equation.
The lines are parallel and distinct → inconsistent system.
Final answer: \(\boxed{\text{Inconsistent system — no solution}}\)
Exercise 11 — level ★★★★☆
\[ \begin{cases} 5x + 2y = 13 \\ 3x + 4y = 19 \end{cases} \]
Answer
\( x = 1 \quad y = 4 \)
Solution
Elimination method
Multiply the first equation by 2 to make the coefficients of \( y \) equal:
\( \begin{cases} 10x + 4y = 26 \\ 3x + 4y = 19 \end{cases} \)
Subtracting: \( 7x = 7 \implies x = 1 \). Then \( y = 4 \).
Verification
\( 5 + 8 = 13 \) and \( 3 + 16 = 19 \)
Final answer: \(\boxed{x = 1 \quad y = 4}\)
Exercise 12 — level ★★★★☆
\[ \begin{cases} 2x - 3y = 5 \\ 5x + 2y = 22 \end{cases} \]
Answer
\( x = 4 \quad y = 1 \)
Solution
Elimination method
Multiply the first by 2 and the second by 3 to eliminate \( y \):
\( \begin{cases} 4x - 6y = 10 \\ 15x + 6y = 66 \end{cases} \)
Adding: \( 19x = 76 \implies x = 4 \). Then \( y = 1 \).
Verification
\( 8 - 3 = 5 \) and \( 20 + 2 = 22 \)
Final answer: \(\boxed{x = 4 \quad y = 1}\)
Exercise 13 — level ★★★★☆
\[ \begin{cases} x + y = 4 \\ x \cdot y = 3 \end{cases} \]
Answer
\( (x,y) = (1,3) \) or \( (3,1) \)
Solution
Combined method
From the first: \( x = 4 - y \). Substituting into the second:
\( (4 - y) \cdot y = 3 \implies 4y - y^2 = 3 \implies y^2 - 4y + 3 = 0 \implies (y-1)(y-3) = 0 \)
\( y = 1 \implies x = 3 \); \( y = 3 \implies x = 1 \).
Final answer: \(\boxed{(x,y)=(1,3)\ \text{or}\ (3,1)}\)
Exercise 14 — level ★★★★☆
\[ \begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ x + y - z = 4 \end{cases} \]
Answer
\( x = 3 \quad y = 2 \quad z = 1 \)
Solution
Elimination by subtraction
Subtracting the second from the first: \( 2y = 4 \implies y = 2 \).
Subtracting the third from the first: \( 2z = 2 \implies z = 1 \).
Then \( x = 6 - y - z = 3 \).
Verification
All three original equations are satisfied.
Final answer: \(\boxed{x = 3 \quad y = 2 \quad z = 1}\)
Exercise 15 — level ★★★★☆
\[ \begin{cases} x + y + z = 9 \\ 2x + y + z = 12 \\ x + 2y + z = 11 \end{cases} \]
Answer
\( x = 3 \quad y = 2 \quad z = 4 \)
Solution
Elimination by subtraction
Second minus first: \( x = 3 \).
Third minus first: \( y = 2 \).
Then \( z = 9 - x - y = 4 \).
Verification
All equations are satisfied.
Final answer: \(\boxed{x = 3 \quad y = 2 \quad z = 4}\)
Exercise 16 — level ★★★★☆
\[ \begin{cases} 2x + y - z = 1 \\ x - y + 2z = 5 \\ 3x + 2y + z = 10 \end{cases} \]
Answer
\( x = 1 \quad y = 2 \quad z = 3 \)
Solution
Reduction of the system
Add the first and second to eliminate \( y \): \( 3x + z = 6 \).
Solve the resulting 2×2 system to find \( x = 1 \), \( z = 3 \), then \( y = 2 \).
Verification
All original equations are satisfied.
Final answer: \(\boxed{x = 1 \quad y = 2 \quad z = 3}\)
Exercise 17 — level ★★★★★
\[ \begin{cases} x - y = 1 \\ x^2 + y^2 = 5 \end{cases} \]
Answer
\( (x,y) = (2,1) \) or \( (-1,-2) \)
Solution
Substitution
\( x = y + 1 \). Substituting: \( (y+1)^2 + y^2 = 5 \implies 2y^2 + 2y - 4 = 0 \implies y^2 + y - 2 = 0 \).
Solutions: \( y = 1 \) (\( x=2 \)) or \( y = -2 \) (\( x=-1 \)).
Final answer: \(\boxed{(x,y)=(2,1)\ \text{or}\ (-1,-2)}\)
Exercise 18 — level ★★★★★
\[ \begin{cases} x^2 + y^2 = 10 \\ x \cdot y = 3 \end{cases} \]
Answer
\( (x,y) \in \{(1,3),(3,1),(-1,-3),(-3,-1)\} \)
Solution
Using algebraic identity
\( (x+y)^2 = x^2 + 2xy + y^2 = 10 + 6 = 16 \implies x+y = \pm 4 \).
Solve the quadratic for each case to find the four pairs.
Final answer: \(\boxed{(x,y) \in \{(1,3),(3,1),(-1,-3),(-3,-1)\}}\)
Exercise 19 — level ★★★★★
\[ \begin{cases} 2x + y + z = 7 \\ x + 2y + z = 8 \\ x + y + 2z = 9 \end{cases} \]
Answer
\( x = 1 \quad y = 2 \quad z = 3 \)
Solution
Sum of the three equations
Adding all: \( 4(x + y + z) = 24 \implies x + y + z = 6 \).
Subtract from each original equation to find \( x=1 \), \( y=2 \), \( z=3 \).
Final answer: \(\boxed{x = 1 \quad y = 2 \quad z = 3}\)
Exercise 20 — level ★★★★★
\[ \begin{cases} x + 2y - z = 0 \\ 2x + y + z = 9 \\ 3x - y + 2z = 13 \end{cases} \]
Answer
\( x = 2 \quad y = 1 \quad z = 4 \)
Solution
Reduction to two equations
Add first and second: \( x + y = 3 \). Substitute \( z = x + 2y \) into the third and solve.
Final answer: \(\boxed{x = 2 \quad y = 1 \quad z = 4}\)
Exercise 21 — level ★★★★☆
\[ \begin{cases} x + 2y + z = 8 \\ 2x + y + 3z = 9 \\ 3x + 4y + 2z = 20 \end{cases} \]
Answer
\( x = 4 \quad y = \frac{11}{5} \quad z = -\frac{2}{5} \)
Solution
Elimination method
Eliminate \( x \) between the equations and solve the resulting 2×2 system.
Final answer: \(\boxed{x = 4 \quad y = \frac{11}{5} \quad z = -\frac{2}{5}}\)
Exercise 22 — level ★★★★★
\[ \begin{cases} 2x + y - z = 5 \\ x + 3y + 2z = 8 \\ 3x + 2y + 4z = 15 \end{cases} \]
Answer
\( x = \frac{67}{25} \quad y = \frac{23}{25} \quad z = \frac{32}{25} \)
Solution
System reduction
Combine elimination and substitution to obtain a 2×2 system.
Final answer: \(\boxed{x = \frac{67}{25} \quad y = \frac{23}{25} \quad z = \frac{32}{25}}\)
Exercise 23 — level ★★★★★
\[ \begin{cases} x + y = 6 \\ 2x + ky = 12 \end{cases} \]
Answer
Depends on the value of \( k \)
Solution
Analysis with parameter
Substituting \( x = 6 - y \): \( (k - 2)y = 0 \).
- If \( k \neq 2 \): unique solution \( x = 6 \), \( y = 0 \)
- If \( k = 2 \): infinitely many solutions (\( x = 6 - t \), \( y = t \))
Final answer: \(\boxed{\text{Determinate if } k \neq 2;\ \text{Indeterminate if } k=2}\)
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 method
\( y = 5 - x \). Substituting: \( x^2 + (5 - x)^2 = 13 \implies x^2 - 5x + 6 = 0 \).
Solutions: \( x=2 \) (\( y=3 \)) and \( x=3 \) (\( y=2 \)).
Final 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
\( x = \frac{49}{12} \quad y = \frac{7}{4} \quad z = \frac{19}{12} \)
Solution
Combined elimination
Progressively eliminate \( y \) and solve the resulting system.
Final answer: \(\boxed{x = \frac{49}{12} \quad y = \frac{7}{4} \quad z = \frac{19}{12}}\)