Systems of Second-Degree Equations in Two Variables

Solver for Systems of Second-Degree Equations

Enter the coefficients of the two equations of the form ${ax^2+bxy}+{cy^2+dx+ey}={f.}$ They can be integers, decimals (such as 0.5) or fractions (such as 1/3). An empty box means 0. The calculator chooses a method from the lesson (substitution, addition, a change of variables or factoring), finds all the real solutions of the system and shows the solution step by step. If none of these methods works, the calculator finds the solutions numerically.

What Is a System of Second-Degree Equations?

A system of two equations in two unknowns is of the second degree if one of its equations is of the second degree and the other is of the first or the second degree. For example, in the system

$$\begin{array}{|l}x-y=7\\x^2-xy-y^2=19\end{array}$$

the first equation is of the first degree and the second is of the second degree (the terms ${x^2,}$ ${xy}$ and ${y^2}$ are of the second degree). Every second-degree equation in two unknowns can be written in the form

A solution of the system is an ordered pair of numbers, written ${(x,\ y),}$ that satisfies both equations. Solving a system means finding all its solutions or showing that it has none. For example, ${(4,\ -3)}$ is a solution of the system above, because ${4-(-3)=7}$ and ${4^2-4\cdot(-3)-(-3)^2}={16+12-9}={19.}$ The pair ${(8,\ 1)}$ satisfies the first equation but not the second (${64-8-1=55\ne 19}$), so it is not a solution.

Two systems are equivalent if they have the same solutions (in particular, if neither of them has any solutions). The methods below replace the given system with a simpler system equivalent to it, or with several simpler systems whose solutions we combine.

On this page: Solver · Substitution · Addition method · Eliminating an unknown · Symmetric systems · Factoring · Homogeneous equations · Word problems · Practice

Systems in Which One Equation Is of the First Degree: The Substitution Method

We can always solve a first-degree equation for one of its unknowns. After the substitution we get an equation in one unknown, most often a quadratic equation. That is why substitution solves every second-degree system in which one of the equations is of the first degree.

1. Use the first-degree equation to express one unknown in terms of the other. It is convenient to choose an unknown whose coefficient is 1 or ${-1.}$
2. Substitute this expression into the second-degree equation to get an equation in one unknown.
3. Solve this equation (usually a quadratic equation).
4. For each root, find the other unknown from the expression in step 1.
5. Write the solutions as pairs ${(x,\ y).}$

In step 4, find the other unknown from the expression you got from the first-degree equation, not from the second-degree equation. Otherwise you may get pairs that are not solutions. For example, in Example 1 below, for ${x=3}$ the equation ${x^2+y^2=25}$ gives ${y=4}$ and ${y=-4,}$ but ${(3,\ -4)}$ is not a solution of the system, because ${3+(-4)\ne 7.}$

Example 1

Solve the system

$$\begin{array}{|l}x+y=7\\x^2+y^2=25\end{array}$$

Solution. From the first equation, ${y=7-x.}$ Substitute into the second equation:

${x^2+(7-x)^2=25}$

$x^2+49-14x+x^2=25$

${2x^2-14x+24}={0\iff x^2-7x+12}={0}$

The roots are ${x_1=3}$ and ${x_2=4}$ (their sum is 7 and their product is 12). From ${y=7-x}$ we get ${y_1=7-3=4}$ and ${y_2=7-4=3.}$

Answer: ${(3,\ 4)}$ and ${(4,\ 3).}$

Example 2

Solve the system

$$\begin{array}{|l}x-y=7\\x^2-xy-y^2=19\end{array}$$

Solution. From the first equation, ${x=7+y.}$ Substitute into the second equation:

${(7+y)^2-(7+y)y-y^2}={19}$

${49+14y+y^2}-{7y-y^2-y^2}={19}$

${-y^2+7y+30}={0\iff y^2-7y-30}={0}$

${D}={(-7)^2-4\cdot 1\cdot(-30)}={49+120}={169,}$ so ${y_{1,2}=\frac{7\pm 13}{2},}$ that is ${y_1=-3}$ and ${y_2=10.}$ From ${x=7+y}$ we get ${x_1=4}$ and ${x_2=17.}$

Check for ${(17,\ 10)\text{:}}$ ${17-10=7}$ and ${17^2-17\cdot 10-10^2}={289-170-100}={19.}$

Answer: ${(4,\ -3)}$ and ${(17,\ 10).}$

Example 3: One Solution or None

a) Solve the system

$$\begin{array}{|l}3x+4y=25\\x^2+y^2=25\end{array}$$

Solution. From the first equation, ${y=\frac{25-3x}{4}.}$ Substitute into the second equation and multiply by 16:

${x^2+\frac{(25-3x)^2}{16}}={25\iff 16x^2+625}-{150x+9x^2}={400}$

${25x^2-150x+225}={0\iff x^2-6x+9}={0\iff (x-3)^2}={0}$

The equation has only one root, ${x=3,}$ so ${y=\frac{25-9}{4}=4.}$ Answer: ${(3,\ 4).}$ The system has one solution.

b) Solve the system

$$\begin{array}{|l}x+y=10\\x^2+y^2=25\end{array}$$

Solution. From the first equation, ${y=10-x.}$ Substitute into the second equation:

${x^2+(10-x)^2}={25\iff 2x^2-20x+75}={0}$

${D}={(-20)^2-4\cdot 2\cdot 75}={400-600}={-200}\lt{0,}$ so the quadratic equation has no real roots. Answer: the system has no solution.

Systems in Which Both Equations Are of the Second Degree: The Elimination (Addition) Method

When both equations are of the second degree, usually neither unknown can be conveniently expressed in terms of the other. Then we look for a way to get a simpler equation. Some of these systems can be solved by addition, or by addition followed by substitution.

We choose the numbers so that adding the equations eliminates the second-degree terms. We get a first-degree equation and continue with substitution. (Subtracting one equation from another is the same as adding it multiplied by ${-1.}$)

Example 4

Solve the system

$$\begin{array}{|l}x^2+y^2+2x=39\\x^2+y^2-y=27\end{array}$$

Solution. Here neither unknown can be conveniently expressed in terms of the other, because both equations contain ${x^2}$ and ${y^2.}$ Notice, however, that if we subtract the second equation from the first, the terms ${x^2}$ and ${y^2}$ cancel:

${(x^2+y^2+2x)}-{(x^2+y^2-y)}={39-27\iff 2x+y}={12}$

We get a system equivalent to the given one:

$$\begin{array}{|l}2x+y=12\\x^2+y^2+2x=39\end{array}$$

We solve it by substitution. From the first equation, ${y=12-2x,}$ and the second equation gives

${x^2+(12-2x)^2+2x}={39\iff 5x^2-46x+105}={0}$

${D}={(-46)^2-4\cdot 5\cdot 105}={2116-2100}={16,}$ so ${x_{1,2}=\frac{46\pm 4}{10},}$ that is ${x_1=5}$ and ${x_2=\frac{21}{5}.}$ Then ${y_1=12-10=2}$ and ${y_2=12-\frac{42}{5}=\frac{18}{5}.}$

Answer: ${(5,\ 2)}$ and ${\left(\frac{21}{5},\ \frac{18}{5}\right),}$ that is ${(4.2,\ 3.6).}$

Example 5

Solve the system

$$\begin{array}{|l}4x^2-2xy+x+y=15\\2x^2-xy+2x-y=9\end{array}$$

Solution. The second-degree terms of the first equation are twice those of the second: ${4x^2-2xy}={2(2x^2-xy).}$ If we multiply the second equation by ${-2}$ and add it to the first, we get a first-degree equation:

$$\begin{array}{|l}4x^2-2xy+x+y=15\\-4x^2+2xy-4x+2y=-18\end{array}$$

$-3x+3y=-3\iff x-y=1$

The system is equivalent to the system

$$\begin{array}{|l}x-y=1\\2x^2-xy+2x-y=9\end{array}$$

From the first equation, ${x=1+y.}$ Substitute into the second equation:

${2(1+y)^2}-{y(1+y)+2(1+y)-y}={9\iff y^2+4y-5}={0}$

The roots are ${y_1=-5}$ and ${y_2=1}$ (their sum is ${-4,}$ and their product is ${-5}$). Then ${x_1=1+(-5)=-4}$ and ${x_2=1+1=2.}$

Answer: ${(-4,\ -5)}$ and ${(2,\ 1).}$

Eliminating One Unknown

Sometimes, by adding the equations multiplied by suitable numbers, we can remove all the terms with one of the unknowns, not only the second-degree terms. We get an equation in the other unknown only. We solve it, and for each of its roots we find the eliminated unknown from one of the equations of the system. By the theorem on adding equations, the new equation and either of the given equations form a system equivalent to the given one. So every pair found in this way is a solution.

This does not contradict the warning given for the substitution method. There, the first-degree equation links ${x}$ and ${y,}$ while here the new equation contains only one of the unknowns. So for each of its roots, every value of the other unknown that satisfies the equation we kept gives a solution.

Example 6

Solve the system

$$\begin{array}{|l}x^2+y^2=13\\x^2-y^2=5\end{array}$$

Solution. Add the two equations: ${2x^2=18,}$ so ${x^2=9}$ and ${x=\pm 3.}$ Subtract the second equation from the first: ${2y^2=8,}$ so ${y^2=4}$ and ${y=\pm 2.}$ The system is equivalent to the system ${x^2=9,}$ ${y^2=4,}$ so each of the two values of ${x}$ can be paired with each of the two values of ${y.}$

Answer: ${(3,\ 2),}$ ${(3,\ -2),}$ ${(-3,\ 2)}$ and ${(-3,\ -2).}$

Example 7

Solve the system

$$\begin{array}{|l}x^2+2y^2-5x+4y=-2\\x^2-3y^2-6y=-2\end{array}$$

Solution. The terms with ${y}$ are ${2y^2+4y}$ in the first equation and ${-3y^2-6y}$ in the second. If we multiply the first equation by 3 and the second by 2 and add them, all the terms with ${y}$ cancel:

$$\begin{array}{|l}3x^2+6y^2-15x+12y=-6\\2x^2-6y^2-12y=-4\end{array}$$

${5x^2-15x}={-10\iff x^2-3x+2}={0}$

The roots are ${x_1=1}$ and ${x_2=2.}$ We substitute each of them into the second equation (it is the simpler one):

• for ${x=1\text{:}}$ ${1-3y^2-6y}={-2\iff y^2+2y-1}={0,}$ so ${y=-1\pm\sqrt{2};}$
• for ${x=2\text{:}}$ ${4-3y^2-6y}={-2\iff y^2+2y-2}={0,}$ so ${y=-1\pm\sqrt{3}.}$

Answer: the system has four solutions, ${(1,\ -1+\sqrt{2}),}$ ${(1,\ -1-\sqrt{2}),}$ ${(2,\ -1+\sqrt{3})}$ and ${(2,\ -1-\sqrt{3}).}$

Symmetric Systems

A system is symmetric if neither of its equations changes when we swap ${x}$ and ${y.}$ For example, the equation ${x^2+xy+y^2=7}$ is symmetric: after the swap we get ${y^2+yx+x^2=7,}$ which is the same equation. Symmetric systems are easy to solve with the substitution

If ${t_1}$ and ${t_2}$ are the roots of ${t^2-ut+v=0,}$ the system has the solutions ${(t_1,\ t_2)}$ and ${(t_2,\ t_1)}$ (when ${t_1=t_2}$ this is a single solution ${(t_1,\ t_1)}$). If this equation has no real roots, that is, if ${u^2-4v\lt 0,}$ the corresponding pair ${u,}$ ${v}$ does not give a solution.

Example 8

Solve the system

$$\begin{array}{|l}x+y=5\\xy=6\end{array}$$

Solution. Here we see at once that ${u=5}$ and ${v=6.}$ The numbers ${x}$ and ${y}$ are the roots of the equation ${t^2-5t+6=0,}$ namely ${t_1=2}$ and ${t_2=3.}$ (The same system can also be solved by substitution: ${y=5-x}$ and ${x(5-x)=6.}$)

Answer: ${(2,\ 3)}$ and ${(3,\ 2).}$

Example 9

Solve the system

$$\begin{array}{|l}x^2+y^2=13\\xy=6\end{array}$$

Solution. Let ${x+y=u}$ and ${xy=v.}$ Since ${x^2+y^2=u^2-2v,}$ the system becomes

$$\begin{array}{|l}u^2-2v=13\\v=6\end{array}$$

Hence ${u^2=25,}$ so ${u=5}$ or ${u=-5.}$

• For ${u=5}$ and ${v=6\text{:}}$ ${t^2-5t+6=0,}$ ${t_1=2,}$ ${t_2=3.}$
• For ${u=-5}$ and ${v=6\text{:}}$ ${t^2+5t+6=0,}$ ${t_1=-3,}$ ${t_2=-2.}$

Answer: ${(2,\ 3),}$ ${(3,\ 2),}$ ${(-2,\ -3)}$ and ${(-3,\ -2).}$

Example 10

Solve the system

$$\begin{array}{|l}x^2+xy+y^2=7\\x+xy+y=5\end{array}$$

Solution. The system is symmetric. Since ${x^2+xy+y^2}={(x^2+y^2)+xy}={u^2-2v+v}={u^2-v,}$ the substitution ${x+y=u,}$ ${xy=v}$ gives

$$\begin{array}{|l}u^2-v=7\\u+v=5\end{array}$$

Add the two equations: ${u^2+u=12,}$ that is ${u^2+u-12=0,}$ so ${u_1=3}$ and ${u_2=-4.}$ Then ${v_1=5-3=2}$ and ${v_2=5-(-4)=9.}$

• For ${u=3}$ and ${v=2\text{:}}$ ${t^2-3t+2=0,}$ ${t_1=1,}$ ${t_2=2.}$ We get ${(1,\ 2)}$ and ${(2,\ 1).}$
• For ${u=-4}$ and ${v=9\text{:}}$ the equation ${t^2+4t+9=0}$ has the discriminant ${16-36=-20\lt 0,}$ so this pair does not give any solutions.

Answer: ${(1,\ 2)}$ and ${(2,\ 1).}$

Factoring

If one of the equations can be written in the form ${A\cdot B=0,}$ where ${A}$ and ${B}$ are first-degree expressions, it holds exactly when ${A=0}$ or ${B=0.}$ Then the system splits into two systems, and in each of them one equation is of the first degree. We solve them by substitution and combine their solutions.

Example 11

Solve the system

$$\begin{array}{|l}x^2-5xy+6y^2=0\\x^2+y^2=10\end{array}$$

Solution. The left side of the first equation factors in the same way as ${t^2-5t+6}={(t-2)(t-3)\text{:}}$ ${x^2-5xy+6y^2}={(x-2y)(x-3y)}$ (check this by expanding the brackets). Hence ${x=2y}$ or ${x=3y.}$

• For ${x=2y}$ the second equation gives ${4y^2+y^2=10,}$ so ${y^2=2}$ and ${y=\pm\sqrt{2}.}$ We get ${(2\sqrt{2},\ \sqrt{2})}$ and ${(-2\sqrt{2},\ -\sqrt{2}).}$
• For ${x=3y}$ the second equation gives ${9y^2+y^2=10,}$ so ${y^2=1}$ and ${y=\pm 1.}$ We get ${(3,\ 1)}$ and ${(-3,\ -1).}$

Answer: ${(2\sqrt{2},\ \sqrt{2}),}$ ${(-2\sqrt{2},\ -\sqrt{2}),}$ ${(3,\ 1)}$ and ${(-3,\ -1).}$

Homogeneous Equations

An equation such as ${x^2-5xy+6y^2=0,}$ in which all the terms are of the second degree and the right side is 0, is called homogeneous. If we do not see at once how to factor it, then for ${y\ne 0}$ we divide by ${y^2}$ and let ${t=\frac{x}{y}.}$ From ${x^2-5xy+6y^2=0}$ we get ${t^2-5t+6=0,}$ whose roots are ${t=2}$ and ${t=3,}$ that is ${x=2y}$ or ${x=3y.}$ We check the case ${y=0}$ separately, because we cannot divide by 0. Here, for ${y=0}$ the equation gives ${x^2=0,}$ that is, the point ${(0,\ 0),}$ which does not satisfy ${x^2+y^2=10.}$ (If the equation has no term with ${x^2,}$ for example ${5xy+6y^2=0,}$ then for ${y=0}$ it holds for every ${x.}$ In that case ${y}$ is a common factor: ${y(5x+6y)=0.}$)

If both equations of the system contain only second-degree terms and constant terms, we multiply them by suitable numbers and add them so that the constant terms cancel. We get a homogeneous equation.

Example 12

Solve the system

$$\begin{array}{|l}x^2+xy=15\\xy+y^2=10\end{array}$$

Solution. Multiply the first equation by 2 and the second by ${-3,}$ and add them:

${2x^2+2xy-3xy-3y^2}={30-30\iff 2x^2-xy-3y^2}={0}$

By the theorem on adding equations, the system is equivalent to the system made up of this homogeneous equation and the second equation. If ${y=0,}$ the homogeneous equation gives ${2x^2=0,}$ that is ${x=0,}$ and ${(0,\ 0)}$ does not satisfy ${xy+y^2=10.}$ For ${y\ne 0}$ we divide by ${y^2}$ and let ${t=\frac{x}{y}\text{:}}$ ${2t^2-t-3=0,}$ so ${t_1=-1}$ and ${t_2=\frac{3}{2}.}$ Hence ${x=-y}$ or ${x=\frac{3y}{2}.}$ Factoring gives the same result: ${2x^2-xy-3y^2}={(x+y)(2x-3y).}$

• For ${x=-y}$ the second equation gives ${-y^2+y^2=10,}$ that is ${0=10,}$ which is false.
• For ${x=\frac{3y}{2}}$ the second equation gives ${\frac{3y^2}{2}+y^2=10,}$ so ${y^2=4}$ and ${y=\pm 2.}$ For ${y=2}$ we get ${x=3,}$ and for ${y=-2}$ we get ${x=-3.}$

Answer: ${(3,\ 2)}$ and ${(-3,\ -2).}$ Check: ${9+6=15}$ and ${6+4=10.}$

Another way. If we add the two equations, we get ${x^2+2xy+y^2=25,}$ that is ${(x+y)^2=25}$ and ${x+y=\pm 5.}$ The first equation can be written as ${x(x+y)=15.}$ For ${x+y=5}$ we get ${5x=15,}$ ${x=3,}$ ${y=2,}$ and for ${x+y=-5}$ we get ${x=-3,}$ ${y=-2.}$

Word Problems

Problem 1. The perimeter of a rectangle is 34 cm and its diagonal is 13 cm. Find the sides of the rectangle.

Solution. Let the sides be ${x}$ cm and ${y}$ cm. Then ${2(x+y)=34,}$ and by the Pythagorean theorem ${x^2+y^2=13^2\text{:}}$

$$\begin{array}{|l}x+y=17\\x^2+y^2=169\end{array}$$

The system is symmetric: ${u=x+y=17}$ and ${u^2-2v=169,}$ that is ${289-2v=169}$ and ${v=xy=60.}$ The numbers ${x}$ and ${y}$ are the roots of ${t^2-17t+60=0\text{:}}$ ${t_1=5}$ and ${t_2=12.}$ Answer: the sides are 5 cm and 12 cm. Check: ${5+12=17}$ and ${25+144=169.}$

Problem 2. The product of the digits of a two-digit number is 24, and the sum of the squares of its digits is 52. What could the number be?

Solution. Let the tens digit be ${x,}$ and the units digit ${y.}$ Then ${xy=24}$ and ${x^2+y^2=52.}$ Hence ${(x+y)^2}={x^2+y^2+2xy}={52+48}={100,}$ and since the digits are positive (their product is 24, so neither of them is 0), ${x+y=10.}$ The digits are the roots of ${t^2-10t+24=0,}$ that is 4 and 6. Answer: the number is 46 or 64.

Problem 3. The difference of two numbers is 3 and their product is 40. Find the numbers.

Solution. Let the numbers be ${x}$ and ${y\text{:}}$

$$\begin{array}{|l}x-y=3\\xy=40\end{array}$$

From the first equation, ${x=y+3.}$ Substitute into the second: ${(y+3)y=40,}$ that is ${y^2+3y-40=0,}$ so ${y_1=-8}$ and ${y_2=5.}$ Then ${x_1=-5}$ and ${x_2=8.}$ Answer: the numbers are 8 and 5, or ${-5}$ and ${-8.}$ Both pairs satisfy the conditions: ${8-5=3,}$ ${8\cdot 5=40}$ and ${-5-(-8)=3,}$ ${(-5)\cdot(-8)=40.}$

Problem 4. The hypotenuse of a right triangle is 10 cm and its area is 24 cm². Find the lengths of the legs.

Solution. Let the legs be ${x}$ cm and ${y}$ cm. Then ${x^2+y^2=100}$ and ${\frac{xy}{2}=24,}$ that is ${xy=48.}$ Hence ${(x+y)^2}={100+96}={196}$ and ${x+y=14}$ (lengths are positive). The numbers ${x}$ and ${y}$ are the roots of ${t^2-14t+48=0,}$ that is 6 and 8. Answer: the legs are 6 cm and 8 cm. Check: ${36+64=100}$ and ${\frac{6\cdot 8}{2}=24.}$

Practice Problems

Try to solve them yourself, then open the answer. You can also check your solution with the calculator at the top of the page.

1. Solve the system $\begin{array}{|l}x+y=8\\xy=15\end{array}$

The numbers ${x}$ and ${y}$ are the roots of ${t^2-8t+15=0,}$ that is 3 and 5. Answer: ${(3,\ 5)}$ and ${(5,\ 3).}$

2. Solve the system $\begin{array}{|l}x-y=2\\x^2+y^2=20\end{array}$

${y=x-2,}$ ${x^2+(x-2)^2}={20\iff x^2-2x-8}={0,}$ ${x_1=-2,}$ ${x_2=4.}$ Answer: ${(-2,\ -4)}$ and ${(4,\ 2).}$

3. Solve the system $\begin{array}{|l}2x+y=5\\x^2+y^2=10\end{array}$

${y=5-2x,}$ ${x^2+(5-2x)^2}={10\iff x^2-4x+3}={0,}$ ${x_1=1,}$ ${x_2=3.}$ Answer: ${(1,\ 3)}$ and ${(3,\ -1).}$

4. Solve the system $\begin{array}{|l}x+2y=7\\xy=6\end{array}$

${x=7-2y,}$ ${(7-2y)y}={6\iff 2y^2-7y+6}={0,}$ ${y_1=\frac{3}{2},}$ ${y_2=2.}$ Answer: ${\left(4,\ \frac{3}{2}\right)}$ and ${(3,\ 2).}$

5. Solve the system $\begin{array}{|l}x^2+y^2=41\\x^2-y^2=9\end{array}$

Add: ${2x^2=50,}$ ${x^2=25.}$ Subtract: ${2y^2=32,}$ ${y^2=16.}$ Answer: ${(5,\ 4),}$ ${(5,\ -4),}$ ${(-5,\ 4)}$ and ${(-5,\ -4).}$

6. Solve the system $\begin{array}{|l}x+y=3\\x^2+y^2=1\end{array}$

${y=3-x,}$ ${x^2+(3-x)^2}={1\iff x^2-3x+4}={0,}$ ${D=9-16=-7\lt 0.}$ Answer: the system has no solution.

7. Solve the system $\begin{array}{|l}x^2+y^2-2x=15\\x^2+y^2+4y=25\end{array}$

Subtract the second equation from the first: ${-2x-4y=-10,}$ that is ${x=5-2y.}$ Substitute into the second: ${(5-2y)^2+y^2+4y}={25\iff 5y^2-16y}={0,}$ ${y_1=0,}$ ${y_2=\frac{16}{5}.}$ Answer: ${(5,\ 0)}$ and ${\left(-\frac{7}{5},\ \frac{16}{5}\right).}$

8. Solve the system $\begin{array}{|l}x^2+y^2=20\\x+y=6\end{array}$

${u=6,}$ ${u^2-2v=20,}$ ${v=8,}$ ${t^2-6t+8=0.}$ Answer: ${(2,\ 4)}$ and ${(4,\ 2).}$

9. Solve the system $\begin{array}{|l}x^2+xy+y^2=13\\x+y=4\end{array}$

${x^2+xy+y^2=u^2-v,}$ so ${16-v=13}$ and ${v=3,}$ ${t^2-4t+3=0.}$ Answer: ${(1,\ 3)}$ and ${(3,\ 1).}$

10. Solve the system $\begin{array}{|l}x^2-3xy+2y^2=0\\x^2+y^2=20\end{array}$

${(x-y)(x-2y)=0.}$ For ${x=y\text{:}}$ ${2y^2=20,}$ ${y=\pm\sqrt{10}.}$ For ${x=2y\text{:}}$ ${5y^2=20,}$ ${y=\pm 2.}$ Answer: ${(\sqrt{10},\ \sqrt{10}),}$ ${(-\sqrt{10},\ -\sqrt{10}),}$ ${(4,\ 2)}$ and ${(-4,\ -2).}$

11. Solve the system $\begin{array}{|l}xy=12\\x^2+y^2=25\end{array}$

$(x+y)^2=25+24=49,$ ${x+y=\pm 7.}$ For ${u=7\text{:}}$ ${t^2-7t+12=0;}$ for ${u=-7\text{:}}$ ${t^2+7t+12=0.}$ Answer: ${(3,\ 4),}$ ${(4,\ 3),}$ ${(-3,\ -4)}$ and ${(-4,\ -3).}$

12. The sum of two numbers is 12 and the sum of their squares is 74. Find the numbers.

${x+y=12,}$ ${x^2+y^2=74,}$ so ${xy=\frac{144-74}{2}=35}$ and ${t^2-12t+35=0.}$ Answer: the numbers are 5 and 7.