First-order equations, or linear equations, are a fundamental component of algebra that serves as the foundation for solving more complex mathematical problems. On the YÖS exam, problems on linear equations appear frequently, both in simple form and as part of systems. Mastery of the methods for solving them is critical for success.
---
Definition: A linear equation in one variable is an equation of the form:$$ax + b = 0$$
where:- $a$ and $b$ are coefficients (numbers, $a \neq 0$)- $x$ is the variable (the unknown)
More general form: $ax + b = cx + d$, which by transposing terms reduces to the form $ax + b = 0$.
Definition: Two equations are called equivalent if they have the same set of solutions.
Properties of equivalence:- If you add (or subtract) the same number or expression to both sides, you obtain an equivalent equation- If you multiply (or divide) both sides by the same nonzero number, you obtain an equivalent equation
To solve the equation $ax + b = 0$ (where $a \neq 0$):
Example 1: Solve $3x + 6 = 0$$$3x = -6$$$$x = -2$$
Check: $3(-2) + 6 = -6 + 6 = 0$ ✓
Example 2: Solve $5x = 20$$$x = \frac{20}{5} = 4$$
Example 3: Solve $-2x + 8 = 0$$$-2x = -8$$$$x = 4$$
Example: Solve $\frac{x}{3} + 5 = 8$
Solution:$$\frac{x}{3} = 8 - 5$$$$\frac{x}{3} = 3$$$$x = 9$$
Example: Solve $\frac{2x + 3}{4} = 5$
Solution:$$2x + 3 = 20$$$$2x = 17$$$$x = 8.5$$
Distributive property:$$a(b + c) = ab + ac$$$$a(b - c) = ab - ac$$
Definition: Like terms are terms containing the same variable raised to the same power.
Rule: When combining like terms, their coefficients are added (or subtracted).
Example 1: Solve $2(x + 3) = 10$
Solution:$$2x + 6 = 10 \quad \text{(expanding brackets)}$$$$2x = 4 \quad \text{(moving 6 to the right-hand side)}$$$$x = 2 \quad \text{(dividing by 2)}$$
Example 2: Solve $3(x - 2) + 5 = 2(x + 1)$
Solution:$$3x - 6 + 5 = 2x + 2 \quad \text{(expanding brackets)}$$$$3x - 1 = 2x + 2 \quad \text{(combining like terms)}$$$$3x - 2x = 2 + 1 \quad \text{(moving variables and numbers)}$$$$x = 3 \quad \text{(simplification)}$$
Check: $3(3 - 2) + 5 = 3(1) + 5 = 8$, $2(3 + 1) = 8$ ✓
Example 3: Solve $5x - 3(x - 2) = 2x + 4$
Solution:$$5x - 3x + 6 = 2x + 4 \quad \text{(expanding brackets)}$$$$2x + 6 = 2x + 4 \quad \text{(combining like terms)}$$$$6 = 4 \quad \text{(moving terms)}$$
This is a contradiction, so the equation has no solutions.
Definition: A parameter is a variable that is regarded as a constant when solving the equation.
An equation with a parameter requires consideration of different cases.
Example 1: Solve $ax = 6$ for $x$ (where $a$ is a parameter)
Solution:- If $a \neq 0$: $x = \frac{6}{a}$- If $a = 0$: the equation takes the form $0 = 6$ (no solutions)
Example 2: Solve $(a - 2)x = 3$ for $x$
Solution:- If $a \neq 2$: $x = \frac{3}{a - 2}$- If $a = 2$: the equation $0x = 3$ has no solutions
Example 3: Solve $ax + 2 = bx + 3$ for $x$
Solution:$$ax - bx = 3 - 2$$$$(a - b)x = 1$$
Definition: A system of two linear equations in two variables has the form:
$$\begin{cases}a_1x + b_1y = c_1 \\a_2x + b_2y = c_2\end{cases}$$
where $a_1, b_1, c_1, a_2, b_2, c_2$ are coefficients.
Algorithm:1. Express one variable in terms of the other from one of the equations2. Substitute the expression into the other equation3. Solve the resulting equation in one variable4. Find the second variable
Example:$$\begin{cases}x + y = 5 \\2x - y = 4\end{cases}$$
From the first equation: $y = 5 - x$
Substituting into the second:$$2x - (5 - x) = 4$$$$2x - 5 + x = 4$$$$3x = 9$$$$x = 3$$
Then: $y = 5 - 3 = 2$
Answer: $(3, 2)$
Algorithm:1. Multiply one or both equations by suitable numbers so that the coefficients of one variable become opposite2. Add (or subtract) the equations3. Solve the resulting equation in one variable4. Substitute the value found into one of the original equations
Example:$$\begin{cases}3x + 2y = 8 \\5x - 2y = 12\end{cases}$$
Adding the equations (the coefficients of $y$ are already opposite):$$3x + 5x = 8 + 12$$$$8x = 20$$$$x = 2.5$$
Substituting into the first equation:$$3(2.5) + 2y = 8$$$$7.5 + 2y = 8$$$$2y = 0.5$$$$y = 0.25$$
Answer: $(2.5; 0.25)$
Description: Each equation represents a straight line in the coordinate plane. The solution of the system is the point of intersection of these lines.
Example:$$\begin{cases}x + y = 3 \\x - y = 1\end{cases}$$
First line: $y = 3 - x$ (passes through the points $(0, 3)$ and $(3, 0)$)Second line: $y = x - 1$ (passes through the points $(0, -1)$ and $(1, 0)$)
Graphically, we find the point of intersection: $(2, 1)$
Check: $2 + 1 = 3$ ✓, $2 - 1 = 1$ ✓
Cramer's Theorem: For the system$$\begin{cases}a_1x + b_1y = c_1 \\a_2x + b_2y = c_2\end{cases}$$
if $\Delta \neq 0$, then:$$x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}$$
where:$$\Delta = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1$$
$$\Delta_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1b_2 - c_2b_1$$
$$\Delta_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1c_2 - a_2c_1$$
Example:$$\begin{cases}2x + 3y = 7 \\4x - y = 5\end{cases}$$
$$\Delta = \begin{vmatrix} 2 & 3 \\ 4 & -1 \end{vmatrix} = 2(-1) - 4(3) = -2 - 12 = -14$$
$$\Delta_x = \begin{vmatrix} 7 & 3 \\ 5 & -1 \end{vmatrix} = 7(-1) - 5(3) = -7 - 15 = -22$$
$$\Delta_y = \begin{vmatrix} 2 & 7 \\ 4 & 5 \end{vmatrix} = 2(5) - 4(7) = 10 - 28 = -18$$
$$x = \frac{-22}{-14} = \frac{11}{7}, \quad y = \frac{-18}{-14} = \frac{9}{7}$$
Answer: $\left(\frac{11}{7}, \frac{9}{7}\right)$
Definition: A system of three linear equations in three variables has the form:
$$\begin{cases}a_1x + b_1y + c_1z = d_1 \\a_2x + b_2y + c_2z = d_2 \\a_3x + b_3y + c_3z = d_3\end{cases}$$
Example:$$\begin{cases}x + y + z = 6 \\2x - y + z = 3 \\x + 2y - z = 5\end{cases}$$
From the first equation: $z = 6 - x - y$
Substituting into the second:$$2x - y + (6 - x - y) = 3$$$$x - 2y + 6 = 3$$$$x - 2y = -3 \quad \text{...(4)}$$
Substituting into the third:$$x + 2y - (6 - x - y) = 5$$$$2x + 3y - 6 = 5$$$$2x + 3y = 11 \quad \text{...(5)}$$
From (4): $x = 2y - 3$
Substituting into (5):$$2(2y - 3) + 3y = 11$$$$4y - 6 + 3y = 11$$$$7y = 17$$$$y = \frac{17}{7}$$
Then: $x = 2 \cdot \frac{17}{7} - 3 = \frac{34}{7} - \frac{21}{7} = \frac{13}{7}$
And: $z = 6 - \frac{13}{7} - \frac{17}{7} = 6 - \frac{30}{7} = \frac{42 - 30}{7} = \frac{12}{7}$
Answer: $\left(\frac{13}{7}, \frac{17}{7}, \frac{12}{7}\right)$
The method consists of eliminating variables from the equations step by step.
Example: The same system$$\begin{cases}x + y + z = 6 \quad \text{...(1)}\\2x - y + z = 3 \quad \text{...(2)}\\x + 2y - z = 5 \quad \text{...(3)}\end{cases}$$
Subtract (1) from (2): $(2) - (1)$:$$x - 2y = -3 \quad \text{...(4)}$$
Subtract (1) from (3): $(3) - (1)$:$$y - 2z = -1 \quad \text{...(5)}$$
Substitute into (5) or use the substitution method as above.
Definition: - Consistent system — has at least one solution- Inconsistent system — has no solutions- Determinate system — has exactly one solution- Indeterminate system — has infinitely many solutions
For the system $\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$:
Example 1 (Determinate):$$\begin{cases}2x + 3y = 7 \\x - y = 1\end{cases}$$$$\Delta = 2(-1) - 1(3) = -5 \neq 0$$Has one solution.
Example 2 (Indeterminate):$$\begin{cases}2x + 4y = 6 \\x + 2y = 3\end{cases}$$The second equation is half of the first. The system has infinitely many solutions: $y = \frac{3 - x}{2}$ for any $x$.
Example 3 (Inconsistent):$$\begin{cases}2x + 3y = 7 \\2x + 3y = 5\end{cases}$$The left-hand sides are the same, but the right-hand sides differ. There are no solutions.
Statement: Solve $3x - 7 = 2x + 5$
Solution:$$3x - 2x = 5 + 7$$$$x = 12$$
Check: $3(12) - 7 = 36 - 7 = 29$, $2(12) + 5 = 29$ ✓
Statement: Solve $\frac{2(x + 3)}{5} = 4$
Solution:$$2(x + 3) = 20$$$$2x + 6 = 20$$$$2x = 14$$$$x = 7$$
Statement: $$\begin{cases}x + 2y = 8 \\3x - y = 1\end{cases}$$
Solution:From the first: $x = 8 - 2y$
Substituting into the second:$$3(8 - 2y) - y = 1$$$$24 - 6y - y = 1$$$$-7y = -23$$$$y = \frac{23}{7}$$
Then: $x = 8 - 2 \cdot \frac{23}{7} = 8 - \frac{46}{7} = \frac{56 - 46}{7} = \frac{10}{7}$
Answer: $\left(\frac{10}{7}, \frac{23}{7}\right)$
Statement: A store sells apples and oranges. 3 apples and 2 oranges cost 130 rubles. 2 apples and 3 oranges cost 120 rubles. Find the cost of one apple and one orange.
Solution:Let an apple cost $x$ rubles and an orange $y$ rubles.
$$\begin{cases}3x + 2y = 130 \\2x + 3y = 120\end{cases}$$
Addition method. Multiply the first by 3, the second by 2:$$\begin{cases}9x + 6y = 390 \\4x + 6y = 240\end{cases}$$
Subtract:$$5x = 150$$$$x = 30$$
Substituting into the first:$$3(30) + 2y = 130$$$$90 + 2y = 130$$$$y = 20$$
Answer: An apple costs 30 rubles, an orange costs 20 rubles.
Statement: For what values of the parameter $a$ does the equation $a(x - 1) = 2$ have the solution $x = 3$?
Solution:Substituting $x = 3$:$$a(3 - 1) = 2$$$$2a = 2$$$$a = 1$$
Check: For $a = 1$: $1(x - 1) = 2 \Rightarrow x = 3$ ✓
❌ Incorrect: $2x + 5 = 10 \Rightarrow 2x = 10$ (forgot to change the sign of 5)
✓ Correct: $2x + 5 = 10 \Rightarrow 2x = 10 - 5 = 5 \Rightarrow x = 2.5$
❌ Incorrect: $3(x - 2) = 3x - 2$
✓ Correct: $3(x - 2) = 3x - 6$
❌ Incorrect: One variable is expressed, but incorrectly substituted into the second equation.
✓ Correct: After expressing a variable, substitute it into the other equation and simplify.
❌ Incorrect: Finding one variable and forgetting to find the second.
✓ Correct: Always find the values of all variables and verify by substitution.
❌ Incorrect: Cramer's rule is used, but the determinants are calculated incorrectly.
✓ Correct: $\Delta = a_1b_2 - a_2b_1$, calculate carefully.
Substitute the value found back into the original equation. This guarantees correctness.
Do not forget the cases when the parameter takes special values (for example, when the denominator becomes zero).
The main mistake in solving equations is sign errors. Double-check.
When solving, move all terms with the variable to the left-hand side and all numbers to the right-hand side. This systematizes the process.
The Gaussian method with matrices is convenient for systems with three or more variables.
On the YÖS exam, time is limited. Practice rapid solving of simple equations to have time for difficult problems.
First-order equations are the main tool in a mathematician's arsenal. From simple equations in one variable to systems with several variables, these methods are applied everywhere. Mastery of the various methods of solution (substitution, addition, Cramer, Gauss) will allow you to solve YÖS problems effectively and apply these skills in more complex branches of mathematics. Remember that the key to success is practice, attention to detail, and verification of answers.
All course topics · Practice problems
Please enable JavaScript to use the practice app.