## Math 9 Chapter 3 Lesson 6: Solve the problem by making a system of equations (continued)

## 1. Summary of theory

### 1.1. Solution method

To solve the problem by making a system of equations, we follow these steps:

**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

### 1.2. Basic math forms

Motion math form

Math form combining geometric quantities

Math form working together as a team, working individually

Math form of flowing water

Math form to find numbers

Math form combines physics, chemistry

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1: **Solve the system of equations (II) by setting the extra unknowns ( \(u = \dfrac{1}{x};v = \dfrac{1}{y}\)) and then answer the given problem.

\(\left( {II} \right)\,\,\left\{ \matrix{{\displaystyle{1 \over x}} = {\displaystyle{3 \over 2}}. {\displaystyle{1 \ over y}} \hfill \cr {\displaystyle{1 \over x}} + {\displaystyle{1 \over y}} = {\displaystyle{1 \over {24}}} \hfill \cr} \right. \)

**Solution guide: **

Set \(u = \dfrac{1}{x};v = \dfrac{1}{y}\), the system (II) becomes:

\(\eqalign{& \left( {II} \right)\,\,\left\{ \matrix{u = {\displaystyle{3 \over 2}}.v \hfill \cr u + v = {\ displaystyle{1 \over {24}}} \hfill \cr} \right \Leftrightarrow \left\{ \matrix{u = {\displaystyle{3 \over 2}}v \hfill \cr {\displaystyle{3 \ over 2}}v + v = {\displaystyle{1 \over {24}}} \hfill \cr} \right \Leftrightarrow \left\{ \matrix{u = {\displaystyle{3 \over 2}}v \hfill \cr {\displaystyle{5 \over 2}}v = {\displaystyle{1 \over {24}}} \hfill \cr} \right \cr & \Leftrightarrow \left\{ \matrix{u = {\displaystyle{3 \over 2}}v \hfill \cr v = {\displaystyle{1 \over {60}}} \hfill \cr} \right \Leftrightarrow \left\{ \matrix{u = {\ displaystyle{1 \over {40}}} \hfill \cr v = {\displaystyle{1 \over {60}}} \hfill \cr} \right. \cr} \)

Then we have:

\(\left\{ \begin{array}{l}

\dfrac{1}{x} = \dfrac{1}{{40}}\\

\dfrac{1}{y} = \dfrac{1}{{60}}

\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}

x = 40\\

y = 60

\end{array} \right.\)

So the number of days it takes for Team A to complete that route alone is 40 days

The number of days it takes for Team B to complete that route alone is 60 days

**Verse 2: **A rectangular field has perimeter \(250m\). Calculate the area of that field, knowing that the length is reduced by 3 times and the width is increased by 2 times, the perimeter remains the same.

__ Solution guide:__ Let the width and length of the field be \(x,y(<0x .)

According to the problem, we have a system of equations: \(\left\{\begin{matrix} x+y=\frac{250}{2}\\ 2x+\frac{y}{3}=\frac{250}{ 2} \end{matrix}\right.\)

Solving the system we get \(\left\{\begin{matrix} x=50\\ y=75 \end{matrix}\right.\)

So the area of the field is \(xy=50.75=3750(m^2)\)

### 2.2. Advanced exercises

**Question 1: **Let’s solve the above problem in another way (let x be the number of jobs done in a day by team A; y is the number of jobs done in a day by team B). What do you think about this solution?

**Solution guide:**

Let x be the number of jobs done in 1 day by team A

y is the number of jobs done in 1 day by team B (x;y>0)

One day both teams can do \(\dfrac {1}{24}\) work so we have the equation:

\(x + y = \dfrac {1}{24}\)

Every day the work of team A is one and a half times that of team B, so we have the equation

\(x=1.5y\)

Therefore, we have a system of equations:

\(\begin{array}{l}

\left\{ \begin{array}{l}

x + y = \dfrac{1}{{24}}\\

x = 1.5y

\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}

x = 1.5y\\

1.5y + y = \dfrac{1}{{24}}

\end{array} \right.\\

\Leftrightarrow \left\{ \begin{array}{l}

x = 1.5y\\

2.5y = \dfrac{1}{{24}}

\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}

y = \dfrac{1}{{60}}\\

x = 1.5.\dfrac{1}{{60}}

\end{array} \right.\\

\Leftrightarrow \left\{ \begin{array}{l}

y = \dfrac{1}{{60}}\\

x = \dfrac{1}{{40}}

\end{array} \right.\left( {tmdk} \right)

\end{array}\)

In 1 day, team A can do \(\dfrac{1}{{40}}\) so team A can do it alone in 40 days

In 1 day, team B can do \(\dfrac{1}{{60}}\) so team B can do it alone in 60 days

Comment:

In this solution, we do not need to set extra unknowns to solve the system of equations.

**Verse 2: **Given a right triangle, know that increasing each side of the right angle by \(2cm\) increases the area by \(17cm^2\). If the right angled sides are reduced by one side \(3cm\) and one side \(1cm\) respectively, the area is reduced by \(11cm^2\). Find the sides of that right triangle.

**Solution guide:**

Let the two sides of the right angle be \(x,y(x\geq y>3)\)

According to the topic: increasing each side of a right angle \(2cm\), the area increases \(17cm^2\), we have the equation:

\(\frac{1}{2}(x+2)(y+2)=\frac{1}{2}xy+17\)

If the right angled sides are reduced by one side \(3cm\), one side \(1cm\), the area is reduced by \(11cm^2\), we have the equation:

\(\frac{1}{2}(x-3)(y-1)=\frac{1}{2}xy-11\)

Solving the system of two equations we have: \(\left\{\begin{matrix} x+y=15\\ x-3y=25 \end{matrix}\right.\) \(\Rightarrow \left\{\ begin{matrix} x=5\\ y=10 \end{matrix}\right.\)

So the lengths of the three sides of the triangle are \(5;10;5\sqrt{5} (cm)\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **A rectangular schoolyard has a perimeter of 340m. Three times the length four times the width is 20m. Calculate the length and width of the school yard.

**Verse 2: **There are several benches in the classroom. If there are three students in each seat, six students will not have a seat. If there are four students in each seat, there is an extra seat. How many chairs and how many students are there in the class?

**Question 3: **Two workers build a wall together in 7 hours and 12 minutes to finish (lime, mortar and bricks are transported by other workers). If the first person works in 5 hours and the second in 6 hours, then both can build \({3 \over 4}\) the wall. How long does it take each person to build the wall alone?

**Question 4: **In a field transplanted 60 hectares of new rice varieties and 40 hectares of old rice varieties. Harvest all 460 tons of rice). What is the yield of each type of rice per hectare, knowing that 3 hectares of new rice can be harvested less than 4 hectares of old rice, which is 1 ton.

### 3.2. Multiple choice exercises

**Question 1:** In the first month, two groups of workers produced 800 machine parts. In the second month, group I exceeded 15%, group II exceeded 20%. Therefore, 945 machine parts have been produced. So in the first month, each group of workers can produce:

A. \(600;400\)

B. \(400;600\)

C. \(300;500\)

D. \(500;300\)

**Verse 2:** A rectangular piece of land has a perimeter of 500m. Find the area of a piece of land knowing that if the length is decreased by 3 times and the width is increased by 2 times, the perimeter remains the same:

A. \(1500(m^2)\)

B. \(3000(m^2)\)

C. \(15000(m^2)\)

D. \(30000(m^2)\)

**Question 3: **Two faucets running together into a tank, after \(4\frac{4}{5}\) is now full. Every hour, the amount of water in faucet I is equal to \(\frac{3}{2}\) the amount of water in faucet II. So how long does it take to fill the tank with each separate faucet?

A. Faucet I is 8 hours, faucet II is 12 hours

B. Faucet I is 12 hours, faucet II is 8 hours

C. Faucet I is 12 hours, faucet II is 16 hours

D. Faucet I is 16 hours, faucet II is 12 hours

**Question 4: **Find a two-digit natural number where the units digit is 3 greater than the tens digit and the product of those two digits is 17 greater than the sum of the two digits.

A. \(36\)

B. \(47\)

C. \(58\)

D. \(69\)

**Question 5: **Find a two-digit number, given that the ones digit is 7, the tens digit is 7, the sum of the squares of the tens and units digit is 65.

A. 81

B. 92

C. 70

D. 69

## 4. Conclusion

Through lessons help students:

- Understand how to solve problems by setting up a system of quadratic equations with two unknowns.
- Students have the skills to solve math problems in textbooks.

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