## Math 9 Review chapter 3: System of quadratic equations with two unknowns

## 1. Summary of theory

### 1.1. First degree equation with two unknowns

A quadratic equation with two unknowns is a line equation of the form \(ax+by=c\)

In which: coefficients a, b, c are given and a; b are not simultaneously equal to 0.

About practice test

A quadratic equation with two unknowns has infinitely many solutions, but they are all dependent on each other

In other words, the solution of the system is written as \(\left\{\begin{matrix} x\epsilon \mathbb{R}\\ y=-\frac{a}{b}x+\frac{c}{ b} \end{matrix}\right (b\neq 0)\)

### 1.2. System of two first degree equations with two unknowns

The system is written in the form:

\(\left\{\begin{matrix} ax+by=c\\ a’x+b’y=c’ \end{matrix}\right.(1)\)

They are equations of the line

The relative positions of the two equations of the line are:

If two lines intersect, then (1) has a unique solution

If two lines coincide, then (1) has infinitely many solutions

If 2 lines are parallel, then (1) has no solution

### 1.3. Solving a system of quadratic equations with two unknowns

**a) Substitution method**

Transform the given system of equations into a new system, where an equation has a hidden

Find that hidden and then deduce the solution of the system

**b) Algebraic addition method**

Multiply both sides of an equation by the appropriate constant so that the coefficients of some unknown are equal or opposite

Add or subtract on the side to cancel a hidden

Find that hidden and then deduce the solution of the system

### 1.4. Solve the problem by making a system of equations

**Step 1:** Set up a system of equations

- Select hide and set conditions for hiding
- Express different quantities in implicit terms
- Based on the problem, make an equation in the form you have learned

**Step 2:** Solve the system of equations

**Step 3: **Compare the results and choose the appropriate solution

## 2. Illustrated exercise

### 2.1. Exercise 1

Which of the following pairs of numbers \((-2;1),(-3;-4),(4;3),(3;0)\) is the solution of the equation \(2x-3y=6\ )

**Solution guide**

Substituting the solutions into the above equation, we get

\(2(-2)-3.1=-7\)

\(2(-3)-3.(-4)=6\)

\(2.4-3.3=1\)

\(2.3-3.0=6\)

So, we only accept two pairs that are \((-3;-4),(3;0)\)

### 2.2. Exercise 2

Without drawing the figure, give the number of solutions of each of the following systems and explain

\(\left\{\begin{matrix} y=2x-5\\ y=3-x \end{matrix}\right.\)

\(\left\{\begin{matrix} y=\frac{1}{2}x+6\\ y=5-2x \end{matrix}\right.\)

\(\left\{\begin{matrix} y=10x+2017\\ y=10x-3 \end{matrix}\right.\)

\(\left\{\begin{matrix} y=2x+1\\ 2x-y+1=0 \end{matrix}\right.\)

**Solution guide**

\(\left\{\begin{matrix} y=2x-5\\ y=3-x \end{matrix}\right.\)

Notice that the above system consists of 2 lines, with different slopes, so they intersect at 1 point, so the system has 1 solution.

\(\left\{\begin{matrix} y=\frac{1}{2}x+6\\ y=5-2x \end{matrix}\right.\)

Similar to the above system, however, there is a special thing that is the product of two slopes is \(\frac{1}{2}.(-2)=-1\) so if we use the geometric method, we see that they are perpendicular to each other.

\(\left\{\begin{matrix} y=10x+2017\\ y=10x-3 \end{matrix}\right.\)

This system consists of two equations with coefficients \(a=a’;b\neq b’\) so they are parallel with no solution.

\(\left\{\begin{matrix} y=2x+1\\ 2x-y+1=0 \end{matrix}\right.\)

Equivalent transformation we get: \(\left\{\begin{matrix} y=2x+1\\y=2x+1 \end{matrix}\right.\)

This system consists of two equations with coefficients \(a=a’;b=b’\) so they overlap and have infinitely many solutions.

### 2.3. Exercise 3

Solve the system by substitution method: \(\left\{\begin{matrix} xy=8\\ 5x+4y=-2 \end{matrix}\right.\); \(\left\{\begin{matrix} 3x+2y=10\\ -x+4y=-9 \end{matrix}\right.\)

**Solution guide**

\(\left\{\begin{matrix} xy=8\\ 5x+4y=-2 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} x=y+8 \\ 5x+4y=-2 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} x=y+8\\ 5(y+8)+4y=-2 \ end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} x=y+8\\ 9y=-42 \end{matrix}\right.\)\(\Leftrightarrow \left\ {\begin{matrix} x=y+8\\ y=-\frac{42}{9} \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} x=\ frac{10}{3}\\ y=-\frac{42}{9} \end{matrix}\right.\)

\(\left\{\begin{matrix} 3x+2y=10\\ -x+4y=-9 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} 3x+ 2y=10 \\ x=9+4y \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} 3(4y+9)+2y=10 \\ x=9+4y \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} 14y=-17 \\ x=9+4y \end{matrix}\right.\)\(\Leftrightarrow \left \{\begin{matrix} y=-\frac{17}{14} \\ x=9+4y \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} y= \frac{29}{7} \\ y=-\frac{17}{14} \end{matrix}\right.\)

### 2.4. Exercise 4

Solve the system by algebraic addition: \(\left\{\begin{matrix} x-3y=10\\ 2x+y=0 \end{matrix}\right.\);\(\left\{\ begin{matrix} 4x-y=8\\ -x+2y=-2 \end{matrix}\right.\)

**Solution guide**

\(\left\{\begin{matrix} x-3y=10\\ 2x+y=0 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} 2x-6y= 20\\ 2x+y=0 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} -7y=20\\ 2x+y=0 \end{matrix}\right. \)\(\Leftrightarrow \left\{\begin{matrix} y=-\frac{20}{7}\\ 2x+y=0 \end{matrix}\right.\)\(\Leftrightarrow \left\ {\begin{matrix} x=\frac{10}{7}\\ y=-\frac{20}{7} \end{matrix}\right.\)

\(\left\{\begin{matrix} 4x-y=8\\ -x+2y=-2 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} 8x- 2y=16\\ -x+2y=-2 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} 7x=14\\ -x+2y=-2 \end{ matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} x=2\\ -x+2y=-2 \end{matrix}\right.\)\(\Leftrightarrow \left\{ \begin{matrix} x=2\\ y=0 \end{matrix}\right.\)

### 2.5. Exercise 5

Find two natural numbers, knowing that their sum is 1006, if the large number is divided by the smaller number, the quotient is 2 and the remainder is 124.

**Solution guide**

Call the two natural numbers to be searched as \(a,b(a,b\epsilon \mathbb{N};a>b>124)\)

According to the topic, we have: \(\left\{\begin{matrix} a+b=1006\\ a=2b+124 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{ matrix} 2b+124+b=1006\\ a=2b+124 \end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} 3b=882\\ a=2b+124 \ end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} b=294\\ a=2b+124 \end{matrix}\right.\)\(\Leftrightarrow \left\{ \begin{matrix} b=294\\ a=712 \end{matrix}\right.\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Solve the following systems of equations:

a) \(\left\{ {\matrix{

{4x + y = – 5} \cr

{3x – 2y = – 12} \cr} } \right.\)

b) \(\left\{ {\matrix{

{x + 3y = 4y – x + 5} \cr

{2x – y = 3x – 2\left( {y + 1} \right)} \cr} } \right.\)

c) \(\left\{ {\matrix{

{3\left( {x + y} \right) + 9 = 2\left( {x – y} \right)} \cr

{2\left( {x + y} \right) = 3\left( {x – y} \right) – 11} \cr} } \right.\)

**Verse 2:** Solve the following systems of equations:

a) \(\left\{ {\matrix{

{\sqrt 3 x – 2\sqrt 2 y = 7} \cr

{\sqrt 2 x + 3\sqrt 3 y = – 2\sqrt 6 } \cr} } \right.\)

b) \(\left\{ {\matrix{

{\left( {\sqrt 2 + 1} \right)x – \left( {2 – \sqrt 3 } \right)y = 2} \cr

{\left( {2 + \sqrt 3 } \right)x + \left( {\sqrt 2 – 1} \right)y = 2} \cr} } \right.\)

**Question 3: **Find the values of \(a\) and \(b\) for the system of equations:

\(\left\{ {\matrix{

{ax + by = 3} \cr

{2ax – 3by = 36} \cr} } \right.\)

whose solution is \((3; -2).\)

**Question 4:** Find a two-digit number knowing that \(2\) times the tens digit is greater than \(5\) times the units digit is \(1\) and the tens digit is divided by the units digit quotient is \(2\) and remainder is \(2.\)

### 3.2. Multiple choice exercises

**Question 1:** Solving the system of equations: \(\left\{\begin{matrix} x=2y+1\\ x-2y+5=0 \end{matrix}\right.\) we get the solution of the system:

A. \((1;0)\)

B. \((3;1)\)

C. \((5;2)\)

D. The system has no solution

**Verse 2:** Calculate the lengths of the two sides of the right angle, knowing that increasing each side by \(3(cm)\) increases the area by \(36(cm^2)\). If you reduce one side by \(2(cm)\) by one side by \(4(cm)\), the area will decrease by \(26(cm^2)\).

A. \((5;7)\)

B. \((8;16)\)

C. \((12;9)\)

D. \((15;18)\)

**Question 3: **Given two line equations \(y=2x-3\) and \(xy=1\). The coordinates of the intersection of those two lines are:

A. \((4;3)\)

B. \((3;4)\)

C. \((-3;4)\)

D. \((4;-3)\)

**Question 4:** The solution of the system \(\left\{\begin{matrix} xy\sqrt{3}=0\\ x\sqrt{3}+2y=3\sqrt{2} \end{matrix}\right.\) is :

A. \(\left\{\begin{matrix} x=-\frac{3\sqrt{6}}{5}\\ y=\frac{3\sqrt{2}}{5} \end{matrix }\right.\)

B. \(\left\{\begin{matrix} x=-\frac{3\sqrt{6}}{5}\\ y=-\frac{3\sqrt{2}}{5} \end{ matrix}\right.\)

C. \(\left\{\begin{matrix} x=\frac{3\sqrt{6}}{5}\\ y=\frac{3\sqrt{2}}{5} \end{matrix} \right.\)

D. \(\left\{\begin{matrix} x=\frac{3\sqrt{6}}{5}\\ y=-\frac{3\sqrt{2}}{5} \end{matrix }\right.\)

**Question 5:** Find a two-digit number, knowing that: If the number is multiplied by the sum of those two digits, the product is 115. If the number is reversed and still multiplied by the sum of those two digits, the product is 60. .

A. 14

B. 41

C. 23

D. 32

## 4. Conclusion

This lesson helps students:

- Systematize knowledge easily.
- Understand the set of solutions of a quadratic equation with two unknowns and its geometric representation.
- Know how to find the general solution formula and draw a line representing the set of solutions of a quadratic equation with two unknowns.

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