A graded collection of 20 practice problems on systems of quadratic equations, designed to build proficiency in solving nonlinear systems by substitution, by equating expressions, and through the use of algebraic identities.
The exercises are arranged in order of increasing difficulty: the sequence begins with straightforward systems and progresses to more involved problems featuring multiple solutions, radicals, and inconsistent systems.
Every solution is worked out in full detail, with careful attention to both the algebraic steps and the verification of the solutions obtained.
Exercise 1 — level ★☆☆☆☆
Solve the system:
\[ \begin{cases} y=x^2,\\ y=4. \end{cases} \]
Answer
\[ S=\{(-2,4),(2,4)\}. \]
Solution
Both equations express the value of \(y\). For the system to be satisfied, the two expressions must be equal.
Setting the right-hand sides equal gives:
\[ x^2=4. \]
We look for all numbers whose square equals \(4\):
\[ x=2 \qquad \text{or} \qquad x=-2. \]
From the second equation we already know that:
\[ y=4. \]
The solution pairs are therefore:
\[ (2,4) \qquad \text{and} \qquad (-2,4). \]
Hence:
\[ S=\{(-2,4),(2,4)\}. \]
Exercise 2 — level ★☆☆☆☆
Solve the system:
\[ \begin{cases} y=x^2,\\ y=x+2. \end{cases} \]
Answer
\[ S=\{(-1,1),(2,4)\}. \]
Solution
Both equations give \(y\) explicitly. Setting the right-hand sides equal:
\[ x^2=x+2. \]
Moving all terms to the left-hand side:
\[ x^2-x-2=0. \]
Factoring the quadratic:
\[ x^2-x-2=(x-2)(x+1). \]
Setting each factor equal to zero:
\[ (x-2)(x+1)=0. \]
Since a product is zero if and only if at least one factor is zero:
\[ x=2 \qquad \text{or} \qquad x=-1. \]
We now compute \(y\) from the second equation:
\[ y=x+2. \]
For \(x=2\):
\[ y=2+2=4. \]
For \(x=-1\):
\[ y=-1+2=1. \]
Therefore:
\[ S=\{(-1,1),(2,4)\}. \]
Exercise 3 — level ★☆☆☆☆
Solve the system:
\[ \begin{cases} x+y=5,\\ xy=6. \end{cases} \]
Answer
\[ S=\{(2,3),(3,2)\}. \]
Solution
From the first equation, we solve for \(y\):
\[ y=5-x. \]
Substituting into the second equation:
\[ x(5-x)=6. \]
Expanding:
\[ 5x-x^2=6. \]
Moving all terms to the left-hand side:
\[ x^2-5x+6=0. \]
Factoring:
\[ x^2-5x+6=(x-2)(x-3). \]
Thus:
\[ x=2 \qquad \text{or} \qquad x=3. \]
We recover the corresponding values of \(y\).
For \(x=2\):
\[ y=5-2=3. \]
For \(x=3\):
\[ y=5-3=2. \]
Therefore:
\[ S=\{(2,3),(3,2)\}. \]
Exercise 4 — level ★☆☆☆☆
Solve the system:
\[ \begin{cases} x-y=1,\\ xy=12. \end{cases} \]
Answer
\[ S=\{(-3,-4),(4,3)\}. \]
Solution
From the first equation:
\[ x=y+1. \]
Substituting into the second equation:
\[ (y+1)y=12. \]
Expanding:
\[ y^2+y=12. \]
Moving all terms to the left-hand side:
\[ y^2+y-12=0. \]
Factoring:
\[ y^2+y-12=(y+4)(y-3). \]
Hence:
\[ y=-4 \qquad \text{or} \qquad y=3. \]
We now find \(x\).
For \(y=-4\):
\[ x=-4+1=-3. \]
For \(y=3\):
\[ x=3+1=4. \]
Therefore:
\[ S=\{(-3,-4),(4,3)\}. \]
Exercise 5 — level ★★☆☆☆
Solve the system:
\[ \begin{cases} x^2+y^2=25,\\ x+y=7. \end{cases} \]
Answer
\[ S=\{(3,4),(4,3)\}. \]
Solution
From the second equation:
\[ y=7-x. \]
Substituting into the first:
\[ x^2+(7-x)^2=25. \]
Expanding the square:
\[ (7-x)^2=x^2-14x+49. \]
The equation becomes:
\[ x^2+x^2-14x+49=25. \]
Collecting like terms:
\[ 2x^2-14x+49=25. \]
Moving all terms to the left-hand side:
\[ 2x^2-14x+24=0. \]
Dividing through by \(2\):
\[ x^2-7x+12=0. \]
Factoring:
\[ x^2-7x+12=(x-3)(x-4). \]
Thus:
\[ x=3 \qquad \text{or} \qquad x=4. \]
We find the corresponding values of \(y\).
For \(x=3\):
\[ y=7-3=4. \]
For \(x=4\):
\[ y=7-4=3. \]
Therefore:
\[ S=\{(3,4),(4,3)\}. \]
Exercise 6 — level ★★☆☆☆
Solve the system:
\[ \begin{cases} x^2+y=12,\\ x+y=6. \end{cases} \]
Answer
\[ S=\{(-2,8),(3,3)\}. \]
Solution
From the second equation, we solve for \(y\):
\[ y=6-x. \]
Substituting into the first equation:
\[ x^2+(6-x)=12. \]
Simplifying:
\[ x^2-x+6=12. \]
Moving all terms to the left-hand side:
\[ x^2-x-6=0. \]
Factoring:
\[ x^2-x-6=(x-3)(x+2). \]
Thus:
\[ x=3 \qquad \text{or} \qquad x=-2. \]
We compute the corresponding values of \(y\).
For \(x=3\):
\[ y=6-3=3. \]
For \(x=-2\):
\[ y=6-(-2)=8. \]
Therefore:
\[ S=\{(-2,8),(3,3)\}. \]
Exercise 7 — level ★★☆☆☆
Solve the system:
\[ \begin{cases} y=x^2-1,\\ y=2x+2. \end{cases} \]
Answer
\[ S=\{(-1,0),(3,8)\}. \]
Solution
Both equations express \(y\) explicitly. Setting the right-hand sides equal:
\[ x^2-1=2x+2. \]
Moving all terms to the left-hand side:
\[ x^2-2x-3=0. \]
Factoring:
\[ x^2-2x-3=(x-3)(x+1). \]
Thus:
\[ x=3 \qquad \text{or} \qquad x=-1. \]
We compute \(y\) from the linear equation:
\[ y=2x+2. \]
For \(x=3\):
\[ y=2\cdot 3+2=8. \]
For \(x=-1\):
\[ y=2\cdot(-1)+2=0. \]
Hence:
\[ S=\{(-1,0),(3,8)\}. \]
Exercise 8 — level ★★☆☆☆
Solve the system:
\[ \begin{cases} x^2+y^2=13,\\ x-y=1. \end{cases} \]
Answer
\[ S=\{(-2,-3),(3,2)\}. \]
Solution
From the second equation:
\[ x=y+1. \]
Substituting into the first equation:
\[ (y+1)^2+y^2=13. \]
Expanding:
\[ y^2+2y+1+y^2=13. \]
Collecting like terms:
\[ 2y^2+2y+1=13. \]
Moving all terms to the left-hand side:
\[ 2y^2+2y-12=0. \]
Dividing through by \(2\):
\[ y^2+y-6=0. \]
Factoring:
\[ y^2+y-6=(y+3)(y-2). \]
Thus:
\[ y=-3 \qquad \text{or} \qquad y=2. \]
We find \(x\) from \(x=y+1\).
For \(y=-3\):
\[ x=-3+1=-2. \]
For \(y=2\):
\[ x=2+1=3. \]
Therefore:
\[ S=\{(-2,-3),(3,2)\}. \]
Exercise 9 — level ★★☆☆☆
Solve the system:
\[ \begin{cases} x^2-y=8,\\ x+y=4. \end{cases} \]
Answer
\[ S=\{(-4,8),(3,1)\}. \]
Solution
From the second equation:
\[ y=4-x. \]
Substituting into the first equation:
\[ x^2-(4-x)=8. \]
Taking care with the minus sign in front of the parenthesis:
\[ x^2-4+x=8. \]
Moving all terms to the left-hand side:
\[ x^2+x-12=0. \]
Factoring:
\[ x^2+x-12=(x+4)(x-3). \]
Thus:
\[ x=-4 \qquad \text{or} \qquad x=3. \]
We find \(y\) from \(y=4-x\).
For \(x=-4\):
\[ y=4-(-4)=8. \]
For \(x=3\):
\[ y=4-3=1. \]
Therefore:
\[ S=\{(-4,8),(3,1)\}. \]
Exercise 10 — level ★★☆☆☆
Solve the system:
\[ \begin{cases} x^2+y^2=10,\\ y=x+2. \end{cases} \]
Answer
\[ S=\{(-3,-1),(1,3)\}. \]
Solution
The second equation already expresses \(y\) in terms of \(x\):
\[ y=x+2. \]
Substituting into the first equation:
\[ x^2+(x+2)^2=10. \]
Expanding the square:
\[ (x+2)^2=x^2+4x+4. \]
The equation becomes:
\[ x^2+x^2+4x+4=10. \]
Collecting like terms:
\[ 2x^2+4x+4=10. \]
Moving all terms to the left-hand side:
\[ 2x^2+4x-6=0. \]
Dividing through by \(2\):
\[ x^2+2x-3=0. \]
Factoring:
\[ x^2+2x-3=(x+3)(x-1). \]
Thus:
\[ x=-3 \qquad \text{or} \qquad x=1. \]
We find \(y\).
For \(x=-3\):
\[ y=-3+2=-1. \]
For \(x=1\):
\[ y=1+2=3. \]
Therefore:
\[ S=\{(-3,-1),(1,3)\}. \]
Exercise 11 — level ★★★☆☆
Solve the system:
\[ \begin{cases} x^2+y^2=20,\\ x-y=2. \end{cases} \]
Answer
\[ S=\{(-2,-4),(4,2)\}. \]
Solution
From the second equation:
\[ x=y+2. \]
Substituting into the first equation:
\[ (y+2)^2+y^2=20. \]
Expanding the square:
\[ (y+2)^2=y^2+4y+4. \]
The equation becomes:
\[ y^2+4y+4+y^2=20. \]
Collecting like terms:
\[ 2y^2+4y+4=20. \]
Moving all terms to the left-hand side:
\[ 2y^2+4y-16=0. \]
Dividing through by \(2\):
\[ y^2+2y-8=0. \]
Factoring:
\[ y^2+2y-8=(y+4)(y-2). \]
Hence:
\[ y=-4 \qquad \text{or} \qquad y=2. \]
We find \(x\) from \(x=y+2\).
For \(y=-4\):
\[ x=-4+2=-2. \]
For \(y=2\):
\[ x=2+2=4. \]
Therefore:
\[ S=\{(-2,-4),(4,2)\}. \]
Exercise 12 — level ★★★☆☆
Solve the system:
\[ \begin{cases} x+y=1,\\ x^2+y^2=13. \end{cases} \]
Answer
\[ S=\{(-2,3),(3,-2)\}. \]
Solution
From the first equation:
\[ y=1-x. \]
Substituting into the second equation:
\[ x^2+(1-x)^2=13. \]
Expanding:
\[ (1-x)^2=x^2-2x+1. \]
The equation becomes:
\[ x^2+x^2-2x+1=13. \]
Collecting like terms:
\[ 2x^2-2x+1=13. \]
Moving all terms to the left-hand side:
\[ 2x^2-2x-12=0. \]
Dividing through by \(2\):
\[ x^2-x-6=0. \]
Factoring:
\[ x^2-x-6=(x-3)(x+2). \]
Hence:
\[ x=3 \qquad \text{or} \qquad x=-2. \]
We find \(y\) from \(y=1-x\).
For \(x=3\):
\[ y=1-3=-2. \]
For \(x=-2\):
\[ y=1-(-2)=3. \]
Therefore:
\[ S=\{(-2,3),(3,-2)\}. \]
Exercise 13 — level ★★★☆☆
Solve the system:
\[ \begin{cases} y=x^2-4x,\\ y=-3. \end{cases} \]
Answer
\[ S=\{(1,-3),(3,-3)\}. \]
Solution
The second equation gives the value of \(y\) directly:
\[ y=-3. \]
Substituting into the first equation:
\[ -3=x^2-4x. \]
Moving all terms to the left-hand side:
\[ x^2-4x+3=0. \]
Factoring:
\[ x^2-4x+3=(x-1)(x-3). \]
Thus:
\[ x=1 \qquad \text{or} \qquad x=3. \]
In both cases the value of \(y\) is:
\[ y=-3. \]
The solution pairs are:
\[ (1,-3) \qquad \text{and} \qquad (3,-3). \]
Therefore:
\[ S=\{(1,-3),(3,-3)\}. \]
Exercise 14 — level ★★★☆☆
Solve the system:
\[ \begin{cases} x^2+y=10,\\ y=x+4. \end{cases} \]
Answer
\[ S=\{(-3,1),(2,6)\}. \]
Solution
The second equation already gives \(y\) in terms of \(x\):
\[ y=x+4. \]
Substituting into the first equation:
\[ x^2+(x+4)=10. \]
This gives:
\[ x^2+x+4=10. \]
Moving all terms to the left-hand side:
\[ x^2+x-6=0. \]
Factoring:
\[ x^2+x-6=(x+3)(x-2). \]
Thus:
\[ x=-3 \qquad \text{or} \qquad x=2. \]
We find the corresponding values of \(y\).
For \(x=-3\):
\[ y=-3+4=1. \]
For \(x=2\):
\[ y=2+4=6. \]
Therefore:
\[ S=\{(-3,1),(2,6)\}. \]
Exercise 15 — level ★★★☆☆
Solve the system:
\[ \begin{cases} x^2+y^2=5,\\ xy=2. \end{cases} \]
Answer
\[ S=\{(-2,-1),(-1,-2),(1,2),(2,1)\}. \]
Solution
The system involves the sum of squares and the product \(xy\). We exploit the identity:
\[ (x+y)^2=x^2+2xy+y^2. \]
Since:
\[ x^2+y^2=5 \qquad \text{and} \qquad xy=2, \]
we obtain:
\[ (x+y)^2=5+2\cdot 2=9. \]
Therefore:
\[ x+y=3 \qquad \text{or} \qquad x+y=-3. \]
We consider the two cases separately.
If:
\[ x+y=3 \qquad \text{and} \qquad xy=2, \]
the two numbers are \(1\) and \(2\), giving the pairs:
\[ (1,2) \qquad \text{and} \qquad (2,1). \]
If instead:
\[ x+y=-3 \qquad \text{and} \qquad xy=2, \]
the two numbers are \(-1\) and \(-2\), giving the pairs:
\[ (-1,-2) \qquad \text{and} \qquad (-2,-1). \]
Therefore:
\[ S=\{(-2,-1),(-1,-2),(1,2),(2,1)\}. \]
Exercise 16 — level ★★★☆☆
Solve the system:
\[ \begin{cases} x+y=4,\\ x^2+y^2=10. \end{cases} \]
Answer
\[ S=\{(1,3),(3,1)\}. \]
Solution
From the first equation:
\[ y=4-x. \]
Substituting into the second equation:
\[ x^2+(4-x)^2=10. \]
Expanding the square:
\[ (4-x)^2=x^2-8x+16. \]
The equation becomes:
\[ x^2+x^2-8x+16=10. \]
Collecting like terms:
\[ 2x^2-8x+16=10. \]
Moving all terms to the left-hand side:
\[ 2x^2-8x+6=0. \]
Dividing through by \(2\):
\[ x^2-4x+3=0. \]
Factoring:
\[ x^2-4x+3=(x-1)(x-3). \]
Therefore:
\[ x=1 \qquad \text{or} \qquad x=3. \]
We find the corresponding values of \(y\) from \(y=4-x\).
For \(x=1\):
\[ y=4-1=3. \]
For \(x=3\):
\[ y=4-3=1. \]
Hence:
\[ S=\{(1,3),(3,1)\}. \]
Exercise 17 — level ★★★★☆
Solve the system:
\[ \begin{cases} x^2+y^2=25,\\ xy=12. \end{cases} \]
Answer
\[ S=\{(-4,-3),(-3,-4),(3,4),(4,3)\}. \]
Solution
The system involves \(x^2+y^2\) and \(xy\). To obtain information about the sum \(x+y\), we use the identity:
\[ (x+y)^2=x^2+2xy+y^2. \]
From the system:
\[ x^2+y^2=25 \qquad \text{and} \qquad xy=12. \]
Therefore:
\[ (x+y)^2=25+2\cdot 12=49. \]
This yields two possibilities:
\[ x+y=7 \qquad \text{or} \qquad x+y=-7. \]
We consider the two cases separately.
If:
\[ x+y=7 \qquad \text{and} \qquad xy=12, \]
the two numbers are \(3\) and \(4\), giving the pairs:
\[ (3,4) \qquad \text{and} \qquad (4,3). \]
If instead:
\[ x+y=-7 \qquad \text{and} \qquad xy=12, \]
the two numbers are \(-3\) and \(-4\), giving the pairs:
\[ (-3,-4) \qquad \text{and} \qquad (-4,-3). \]
Therefore:
\[ S=\{(-4,-3),(-3,-4),(3,4),(4,3)\}. \]
Exercise 18 — level ★★★★☆
Solve the system:
\[ \begin{cases} x^2+y^2=8,\\ x-y=2. \end{cases} \]
Answer
\[ S= \left\{ \left(1+\sqrt{3},-1+\sqrt{3}\right), \left(1-\sqrt{3},-1-\sqrt{3}\right) \right\}. \]
Solution
From the second equation:
\[ x=y+2. \]
Substituting into the first equation:
\[ (y+2)^2+y^2=8. \]
Expanding the square:
\[ (y+2)^2=y^2+4y+4. \]
The equation becomes:
\[ y^2+4y+4+y^2=8. \]
Collecting like terms:
\[ 2y^2+4y+4=8. \]
Moving all terms to the left-hand side:
\[ 2y^2+4y-4=0. \]
Dividing through by \(2\):
\[ y^2+2y-2=0. \]
This quadratic does not factor over the integers, so we apply the quadratic formula:
\[ y=\frac{-2\pm\sqrt{2^2-4\cdot 1\cdot(-2)}}{2}. \]
Computing the discriminant:
\[ \Delta=4+8=12. \]
Thus:
\[ y=\frac{-2\pm\sqrt{12}}{2}. \]
Since \(\sqrt{12}=2\sqrt{3}\), this simplifies to:
\[ y=\frac{-2\pm 2\sqrt{3}}{2} = -1\pm\sqrt{3}. \]
We now find \(x\) from \(x=y+2\).
For \(y=-1+\sqrt{3}\):
\[ x=-1+\sqrt{3}+2=1+\sqrt{3}. \]
For \(y=-1-\sqrt{3}\):
\[ x=-1-\sqrt{3}+2=1-\sqrt{3}. \]
Therefore:
\[ S= \left\{ \left(1+\sqrt{3},-1+\sqrt{3}\right), \left(1-\sqrt{3},-1-\sqrt{3}\right) \right\}. \]
Exercise 19 — level ★★★★☆
Solve the system:
\[ \begin{cases} x^2+y^2=1,\\ x+y=3. \end{cases} \]
Answer
\[ S=\varnothing. \]
Solution
From the second equation:
\[ y=3-x. \]
Substituting into the first equation:
\[ x^2+(3-x)^2=1. \]
Expanding the square:
\[ (3-x)^2=x^2-6x+9. \]
The equation becomes:
\[ x^2+x^2-6x+9=1. \]
Collecting like terms:
\[ 2x^2-6x+9=1. \]
Moving all terms to the left-hand side:
\[ 2x^2-6x+8=0. \]
Dividing through by \(2\):
\[ x^2-3x+4=0. \]
Computing the discriminant:
\[ \Delta=(-3)^2-4\cdot 1\cdot 4=9-16=-7. \]
Since \(\Delta<0\), the equation has no real solutions.
Consequently, the system is inconsistent over the reals:
\[ S=\varnothing. \]
Exercise 20 — level ★★★★★
Solve the system:
\[ \begin{cases} x^2+y^2=13,\\ x^2-y^2=5. \end{cases} \]
Answer
\[ S=\{(-3,-2),(-3,2),(3,-2),(3,2)\}. \]
Solution
The system involves \(x^2+y^2\) and \(x^2-y^2\). The efficient strategy is to add and subtract the two equations, isolating \(x^2\) and \(y^2\) separately.
Adding the equations:
\[ (x^2+y^2)+(x^2-y^2)=13+5. \]
The \(y^2\) terms cancel on the left:
\[ 2x^2=18. \]
Thus:
\[ x^2=9. \]
Hence:
\[ x=3 \qquad \text{or} \qquad x=-3. \]
Subtracting the second equation from the first:
\[ (x^2+y^2)-(x^2-y^2)=13-5. \]
The \(x^2\) terms cancel on the left:
\[ 2y^2=8. \]
Thus:
\[ y^2=4. \]
Hence:
\[ y=2 \qquad \text{or} \qquad y=-2. \]
Since both equations involve only \(x^2\) and \(y^2\), all sign combinations are valid.
Therefore:
\[ S=\{(-3,-2),(-3,2),(3,-2),(3,2)\}. \]