---

# Systems of Linear Equations in Two Variables

## General Method of Solution

1. **Express one variable** in terms of the other from one of the equations.
2. **Substitute** this expression into the other equation, obtaining an equation in one variable.
3. **Solve** this equation.
4. **Back-substitute** the result into one of the original equations to find the other variable.

---

## Remarks

* This method is called the **substitution method**.
* It often works well, but it may also lead to difficulties:

* sometimes the algebra becomes complicated,
* in some cases the method is not applicable,
* and sometimes it is not the most efficient choice.
* Systems of equations can also have **no solution** or **infinitely many solutions**.

---

## Aim of This Material

In the following sections we present the details and the most important types of systems of equations **through worked examples**. Each example illustrates a typical situation and highlights the possible pitfalls.

---

# Solving Systems of Linear Equations

---

## Problem 1

Solve the system:

$$
\begin{cases}
x - y = 2 \\
3x - 2y = 9
\end{cases}
$$

**Step 1. Express $x$ in terms of $y$:**

$$
x = y + 2
$$

**Step 2. Substitute into the second equation:**

$$
3(y+2) - 2y = 9 \quad \Rightarrow \quad y + 6 = 9 \quad \Rightarrow \quad y = 3
$$

**Step 3. Solve for $x$:**

$$
x = y + 2 = 3 + 2 = 5
$$

**Answer:**

$$
x = 5, \; y = 3
$$

---

## Problem 2

Solve the system:

$$
\begin{cases}
2x + y = 5 \\
x - 3y = -1
\end{cases}
$$

**Step 1. Express $y$ from the first equation:**

$$
y = 5 - 2x
$$

**Step 2. Substitute into the second equation:**

$$
x - 3(5 - 2x) = -1
$$

$$
x - 15 + 6x = -1
$$

$$
7x - 15 = -1 \quad \Rightarrow \quad 7x = 14 \quad \Rightarrow \quad x = 2
$$

**Step 3. Solve for $y$:**

$$
y = 5 - 2(2) = 1
$$

**Answer:**

$$
x = 2, \; y = 1
$$

---

## Problem 3

Solve the system:

$$
\begin{cases}
x + \tfrac{3y}{4} = 13 \\
\tfrac{x}{2} - \tfrac{y}{3} = 6
\end{cases}
$$

**Step 1. Multiply the second equation by 6 to clear denominators:**

$$
\frac{x}{2} - \frac{y}{3} = 6 \quad \Rightarrow \quad 3x - 2y = 36
$$

**Step 2. Multiply the first equation by 4:**

$$
x + \frac{3y}{4} = 13 \quad \Rightarrow \quad 4x + 3y = 52
$$

So the system is:

$$
\begin{cases}
4x + 3y = 52 \\
3x - 2y = 36
\end{cases}
$$

**Step 3. Eliminate one variable:**
Multiply the second equation by 3:

$$
9x - 6y = 108
$$

Multiply the first equation by 2:

$$
8x + 6y = 104
$$

Add the resulting equations together:

$$
9x + 8x = 108 + 104$$

$$
17x = 212 \quad \Rightarrow \quad x = 212/17
$$

**Step 4. Solve for $y$:**
From (1):

$$
4x + 3y = 52
$$

$$
3y = 52 - 4x
$$

$$
3y = 52 - 4 \cdot \frac{212}{17}
$$

$$
3y = 52 - \frac{848}{17}
$$

$$
3y = \frac{884}{17} - \frac{848}{17} = \frac{36}{17}
$$

$$
y = \frac{12}{17}
$$

**Answer:**

$$
x = \tfrac{212}{17}, \; y = \tfrac{12}{17}
$$

---

## Problem 4

Solve the system:

$$
\begin{cases}
x + y + 5(x - y) = 2 \\
3(x + y) + 6(x - y) = 6
\end{cases}
$$

**Step 1. Simplify each equation.**
First:

$$
x + y + 5x - 5y = 2 \quad \Rightarrow \quad 6x - 4y = 2
$$

Second:

$$
3x + 3y + 6x - 6y = 6 \quad \Rightarrow \quad 9x - 3y = 6
$$

So the system is:

$$
\begin{cases}
6x - 4y = 2 \\
9x - 3y = 6
\end{cases}
$$

**Step 2. Simplify the equations.**
Divide the first equation by 2:

$$
3x - 2y = 1
$$

Divide the second equation by 3:

$$
3x - y = 2
$$

So:

$$
\begin{cases}
3x - 2y = 1 \\
3x - y = 2
\end{cases}
$$

**Step 3. Eliminate $3x$.**
Subtract the first equation from the second:

$$
(3x - y) - (3x - 2y) = 2 - 1
$$

$$
-y + 2y = 1
$$

$$
y = 1
$$

**Step 4. Find $x$.**
From $3x - y = 2$:

$$
3x - 1 = 2 \quad \Rightarrow \quad 3x = 3 \quad \Rightarrow \quad x = 1
$$

**Answer:**

$$
x = 1, \; y = 1
$$

---

# Systems with Fractions & Substitutions

## Problem 5

Solve the system using substitition:

$$
\begin{cases}
x+\dfrac{5}{y}=15\\[2mm]
2x-\dfrac{25}{y}=23
\end{cases}
$$

**Idea.** Let $t=\dfrac{1}{y}$. Then

$$
\begin{cases}
x+5t=15\\
2x-25t=23
\end{cases}
$$

**Solve (linear in $x,t$).** From the first: $x=15-5t$.
Substitute into the second:

$$
2(15-5t)-25t=23 \;\Rightarrow\; 30-10t-25t=23 \;\Rightarrow\; -35t=-7 \;\Rightarrow\; t=\tfrac15.
$$

Hence $y=\tfrac{1}{t}=5$, and then $x=15-5\cdot\tfrac15=15-1=14$.

**Answer:** $x=14,\; y=5$.
(Quick check: $2\cdot14-25/5=28-5=23$ ✓)

---

## Problem 6

Solve the system using substitution:

$$
\begin{cases}
\dfrac{1}{x}-\dfrac{1}{y}=-1\\[2mm]
\dfrac{2}{x}-\dfrac{3}{y}=1
\end{cases}
$$

**Idea.** Let $a=\dfrac{1}{x}$ and $b=\dfrac{1}{y}$. Then

$$
\begin{cases}
a-b=-1\\
2a-3b=1
\end{cases}
$$

From the first: $a=b-1$. Substitute:

$$
2(b-1)-3b=1 \;\Rightarrow\; 2b-2-3b=1 \;\Rightarrow\; -b=3 \;\Rightarrow\; b=-3.
$$

Thus $a=b-1=-4$. Hence $x=\dfrac{1}{a}=-\tfrac14$, $y=\dfrac{1}{b}=-\tfrac13$.

**Answer:** $x=-\tfrac14,\; y=-\tfrac13$.
(Check: $-4-(-3)=-1$; $-8-(-9)=1$ ✓)

---

## Problem 7

Solve the system:

$$
\begin{cases}
\dfrac{2}{x+4}-\dfrac{3}{\,y-1\,}=5\\[2mm]
\dfrac{5}{x+4}-10=\dfrac{5}{\,y-1\,}
\end{cases}
$$

**Idea.** Let $u=\dfrac{1}{x+4}$ and $v=\dfrac{1}{y-1}$. Then

$$
\begin{cases}
2u-3v=5\\
5u-10=5v \;\;\Longleftrightarrow\;\; u-2=v
\end{cases}
$$

From $u-2=v$ we get $u=v+2$. Substitute into $2u-3v=5$:

$$
2(v+2)-3v=5 \;\Rightarrow\; 2v+4-3v=5 \;\Rightarrow\; -v=1 \;\Rightarrow\; v=-1,
$$

so $u=v+2=1$.

Back-substitute: $x+4=\dfrac{1}{u}=1 \Rightarrow x=-3$; $y-1=\dfrac{1}{v}=-1 \Rightarrow y=0$.

**Answer:** $x=-3,\; y=0$.
(Checks: $2/1 - 3/(-1)=2-(-3)=5$; $5/1-10=5/(-1)\Rightarrow -5=-5$ ✓)

---

## Problem 8

Solve the system:

$$
\begin{cases}
\dfrac{x}{y}+y=-3\\[2mm]
\dfrac{2x}{y}-3y=-1
\end{cases}
$$

**Idea.** Let $z=\dfrac{x}{y}$. Then

$$
\begin{cases}
z+y=-3\\
2z-3y=-1
\end{cases}
$$

From the first: $z=-3-y$. Substitute:

$$
2(-3-y)-3y=-1 \;\Rightarrow\; -6-2y-3y=-1 \;\Rightarrow\; -5y=5 \;\Rightarrow\; y=-1.
$$

Then $z=-3-(-1)=-2$. Since $z=\dfrac{x}{y}$, we get $x=zy=(-2)(-1)=2$.

**Answer:** $x=2,\; y=-1$.
(Checks: $2/(-1)+(-1)=-2-1=-3$; $2\cdot2/(-1)-3(-1)=-4+3=-1$ ✓)

---

# Nonlinear Systems of Equations

## Problem 9

**Solve the system (here, we give a solution based on the so called Vieta's formulas, but you can find other solutions too):**

$$
\begin{cases}
x + y = 5,\\
xy = 6.
\end{cases}
$$

**Idea.** Use Vieta's formulas and treat $x$ and $y$ as the two roots of a quadratic with sum $5$ and product $6$.
The quadratic is $t^2 - 5t + 6 = 0$.

**Solve the quadratic:** $(t-2)(t-3)=0 \Rightarrow t=2$ or $t=3$.
Thus the two ordered pairs are permutations of $(2,3)$.

**Answer:** $x_1=2,\, y_1=3,\; x_2=3,\, y_2=2.$

---

## Problem 10

**Solve the system:**

$$
\begin{cases}
x^2 - y = 2,\\
x - y = 2.
\end{cases}
$$

From the second equation: $y = x - 2$. Substitute into the first:

$$
x^2 - (x - 2) = 2 \;\Rightarrow\; x^2 - x + 2 = 2 \;\Rightarrow\; x^2 - x = 0
$$

$$
x(x-1)=0 \;\Rightarrow\; x=0 \text{ or } x=1.
$$

Then $y = x-2$ gives $y=-2$ or $y=-1$.

**Answer:**

* If $x=0$, then $y=-2$.
* If $x=1$, then $y=-1$.
So $x_1=0,\, y_1=-2,\; x_2=1,\, y_2=-1.$

---

## Problem 11

**Solve the system:**

$$
\begin{cases}
x^2 + xy = 35,\\
y^2 + xy = 14.
\end{cases}
$$

Note that $x^2+xy = x(x+y)=35$ and $y^2+xy = y(x+y)=14$.

Let $S=x+y$.

The case $S=0$ leads to the following contradiction: $x(x+y)=35 \;\Rightarrow\; 0=35$.

Thus, $S\neq 0$ and we can divide by $S$. Dividing the equations by $S =x+y$ gives $x=\frac{35}{S}$ and $y=\frac{14}{S}$. Hence

$$
S = x+y = \frac{35+14}{S} = \frac{49}{S} \;\Rightarrow\; S^2=49 \;\Rightarrow\; S=\pm 7.
$$

* If $S=7$: $x=5,\; y=2$.
* If $S=-7$: $x=-5,\; y=-2$.

**Answer:** $x_1=5,\, y_1=2,\; x_2=-5,\, y_2=-2.$

---

## Problem 12

**Solve the system:**

$$
\begin{cases}
x^2 - y^2 = 2(x+y),\\
x^2 + y^2 = 5(x - y).
\end{cases}
$$

From the first equation:

$$
(x-y)(x+y) = 2(x+y) \;\Rightarrow\; (x+y)\big[(x-y)-2\big]=0.
$$

So we have two cases:

### Case 1: $x+y=0$

Plug $y=-x$ into the second equation:

$$
x^2 + x^2 = 5\big(x - (-x)\big) \;\Rightarrow\; 2x^2 = 10x
$$

$$
2x(x-5)=0 \;\Rightarrow\; x=0 \text{ or } x=5.
$$

Thus $(x,y)=(0,0)$ or $(5,-5)$.

### Case 2: $x-y=2$

Plug $y=x-2$ into the second equation:

$$
x^2 + (x-2)^2 = 5\cdot 2 = 10
$$

$$
x^2 + x^2 - 4x + 4 = 10 \;\Rightarrow\; 2x^2 - 4x - 6 = 0
$$

$$
x^2 - 2x - 3 = 0 \;\Rightarrow\; (x-3)(x+1)=0 \Rightarrow x=3 \text{ or } x=-1.
$$

Hence $(x,y)=(3,1)$ or $(-1,-3)$.

**Answer:** $x_1=0,\, y_1=0;\; x_2=5,\, y_2=-5;\; x_3=3,\, y_3=1;\; x_4=-1,\, y_4=-3.$

---

## Problem 13

**Solve the system:**

$$
\begin{cases}
|x| + y = 3,\\
|x-1| = y + 2.
\end{cases}
$$

### Step 1. Eliminate $y$

From the first equation:

$$
y = 3 - |x|.
$$

Substitute into the second:

$$
|x-1| = (3 - |x|) + 2 = 5 - |x|.
$$

(Here both sides are $\ge 0$, so this is consistent only if $|x|\le 5$, which will hold for our solutions.)

### Step 2. Case analysis on $x$

**Case 1: $x \ge 1$**
$|x|=x,\; |x-1|=x-1$.

$$
x-1 = 5 - x \;\Rightarrow\; 2x=6 \;\Rightarrow\; x=3.
$$

Then $y = 3 - |x| = 3 - 3 = 0$.
**Check:** $|x-1|=|2|=2 = y+2$ ✓

**Case 2: $0 \le x < 1$**
$|x|=x,\; |x-1|=1-x$.

$$
1 - x = 5 - x \;\Rightarrow\; 1=5 \quad \text{(impossible)}.
$$

No solution in this interval.

**Case 3: $x < 0$**
$|x|=-x,\; |x-1|=1-x$.

$$
1 - x = 5 - (-x) = 5 + x \;\Rightarrow\; -2x = 4 \;\Rightarrow\; x = -2.
$$

Then $y = 3 - |x| = 3 - 2 = 1$.
**Check:** $|x-1|=|-3|=3 = y+2$ ✓

### Answer:

$$
x_1 = 3,\; y_1 = 0;\quad x_2 = -2,\; y_2 = 1.
$$

---

# Word problems

## Problem 14

At a market, a customer paid 3050 Ft for 4 kg of potatoes and 2 kg of onions. The next customer paid 4425 Ft for 3 kg of potatoes and 6 kg of onions. How much does 1 kg of potatoes cost, and how much does 1 kg of onions cost?

### Step 1. Translate the problem into equations

Let $x$ be the price of 1 kg of potatoes, and $y$ be the price of 1 kg of onions.

* From the first customer: $4x + 2y = 3050$.
* From the second customer: $3x + 6y = 4425$.

### Step 2. Simplify the equations

$$
4x + 2y = 3050 \;\;\Rightarrow\;\; 2x + y = 1525,
$$

$$
3x + 6y = 4425 \;\;\Rightarrow\;\; x + 2y = 1475.
$$

### Step 3. Solve the system

From the second equation:

$$
x = 1475 - 2y.
$$

Substitute into the first:

$$
2(1475 - 2y) + y = 1525,
$$

$$
2950 - 4y + y = 1525,
$$

$$
2950 - 3y = 1525,
$$

$$
-3y = -1425 \;\;\Rightarrow\;\; y = 475.
$$

Then

$$
x = 1475 - 2(475) = 1475 - 950 = 525.
$$

### Answer:

$$
x = 525 \;\text{Ft/kg (potatoes)},\quad y = 475 \;\text{Ft/kg (onions)}.
$$

One kilogram of potatoes costs 525 Ft, and one kilogram of onions costs 475 Ft.

---
## Problem 15

A cinema sold 220 tickets in total for a certain show, consisting of adult tickets at 1800 Ft each and student tickets at 1200 Ft each. The total income from ticket sales was 336,000 Ft. How many adult tickets and how many student tickets were sold?

### Step 1. Translate the problem into equations

Let $x$ be the number of adult tickets, and $y$ be the number of student tickets.

* The total number of tickets: $x + y = 220$.
* The total income: $1800x + 1200y = 336000$.

### Step 2. Simplify the equations

$$
x + y = 220,
$$

$$
1800x + 1200y = 336000 \;\;\Rightarrow\;\; 3x + 2y = 560 \quad \text{(dividing through by 600)}.
$$

### Step 3. Solve the system

From the first equation:

$$
y = 220 - x.
$$

Substitute into the second:

$$
3x + 2(220 - x) = 560,
$$

$$
3x + 440 - 2x = 560,
$$

$$
x + 440 = 560,
$$

$$
x = 120.
$$

Then

$$
y = 220 - 120 = 100.
$$

### Answer:

120 adult tickets and 100 student tickets were sold.

---

## Problem 16

In the rectangular coordinate system, the graph of a linear function passes through the points $(-7, 9)$ and $(0, -1)$. Determine the rule of the function in the form $y = mx + b$.

### Step 1. Set up equations

The function has the form $y = mx + b$.

* Substituting $(-7, 9)$:

$$
9 = m(-7) + b \;\;\Rightarrow\;\; -7m + b = 9.
$$

* Substituting $(0, -1)$:

$$
-1 = m(0) + b \;\;\Rightarrow\;\; b = -1.
$$

### Step 2. Solve the system

From the second equation, $b = -1$.
Substitute into the first:

$$
-7m + (-1) = 9 \;\;\Rightarrow\;\; -7m - 1 = 9,
$$

$$
-7m = 10 \;\;\Rightarrow\;\; m = -\tfrac{10}{7}.
$$

### Step 3. Write the function

$$
y = -\tfrac{10}{7}x - 1.
$$

### Answer:

The rule of the function is $y = -\tfrac{10}{7}x - 1$.