## Math 9 Chapter 3 Lesson 3: Solve system of equations by substitution method

## 1. Summary of theory

The substitution rule is used to transform a system of equations into an equivalent system of equations. The substitution rule consists of the following two steps:

- Step 1: From an equation of the given system (as the first equation), we represent one unknown in terms of the other and then substitute the second equation to get a new equation (only one unknown remains).
- Step 2: Using the new equation to replace one of the two equations of the system, we get a new system of equations equivalent to the original system.

### 1.2. Use the substitution rule to solve the system of equations

- Step 1: Use the substitution rule to transform the given system of equations to get a new equivalent system of equations, in which there is an equation of one unknown.
- Step 2: Solve the equation of that one unknown, then find the remaining unknown, then deduce the solution of the given system.

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1:**Solve the following system of equations using the substitution method \(\left\{\begin{matrix} x-2y=1\\ x+y=1 \end{matrix}\right.\)

**Solution guide**

We have \(\left\{\begin{matrix} x-2y=1\\ x+y=1 \end{matrix}\right.<=>\left\{\begin{matrix} x=2y+1 \\ x+y=1 \end{matrix}\right.\)\(<=>\left\{\begin{matrix} x=2y+1\\ 2y+1+y=1 \end{matrix} \right.\)

\(<=>\left\{\begin{matrix} x=2y+1\\ 3y=0 \end{matrix}\right.\) \(<=>\left\{\begin{matrix} x= 1\\ y=0 \end{matrix}\right.\)

**Verse 2:** Solve the following system of equations by substitution \(\left\{\begin{matrix} -x+2y=1\\ 2x-4y=-2 \end{matrix}\right.\)

**Solution guide**

We have \(\left\{\begin{matrix} -x+2y=1\\ 2x-4y=-2 \end{matrix}\right.<=>\left\{\begin{matrix} x=2y -1\\ 2x-4y=-2 \end{matrix}\right.\)

\(<=>\left\{\begin{matrix} x=2y-1\\ 2(2y-1)-4y=-2 \end{matrix}\right.\) \(<=>\left\ {\begin{matrix} x=2y-1\\ 0y=0 \end{matrix}\right.\)

\(<=>\left\{\begin{matrix} x=2y-1\\ y \in \mathbb{R} \end{matrix}\right.\)

**Question 3: **Prove that the following system of equations has no solution \(\left\{\begin{matrix} x-3y=2\\ -3x+9y=0 \end{matrix}\right.\)

**Solution guide**

We have \(\left\{\begin{matrix} x-3y=2\\ -3x+9y=0 \end{matrix}\right.<=>\left\{\begin{matrix} x=3y+ 2\\ -3x+9y=0 \end{matrix}\right.\)

\(<=>\left\{\begin{matrix} x=3y+2\\ -3(3y+2)+9y=0 \end{matrix}\right.\)

\(<=>\left\{\begin{matrix} x=3y+2\\ 0x=6 \end{matrix}\right.\).

Since the equation \(0x=6\) has no solution, the given system has no solution

### 2.2. Advanced exercises

**Question 1:** Given a system of equations with the parameter a: \(\left\{\begin{matrix} (a+1)xy=a+1\\ x+(a-1)y=2 \end{matrix}\right.\ ).

Solve and argue this system.

**Solution guide**

We have \(\left\{\begin{matrix} (a+1)xy=a+1\\ x+(a-1)y=2 \end{matrix}\right.<=>\left\{\ begin{matrix} y=(a+1)x-(a+1)\\ x+(a-1)y=2 \end{matrix}\right.\)

\(<=>\left\{\begin{matrix} y=(a+1)x-(a+1)\\ x+(a-1)[(a+1)x-(a+1)]=2 \end{matrix}\right.\)

\(<=> \left\{\begin{matrix} y=(a+1)x-(a+1)\\ a^2x=a^2+1 \end{matrix}\right.\)

If \(a \neq 0\) then the equivalent system \(\left\{\begin{matrix} y=(a+1)x-(a+1)\\ x=\frac{a^2+1 }{a^2} \end{matrix}\right <=>\left\{\begin{matrix} y=\frac{a+1}{a^2}\\ x=\frac{a^2 +1}{a^2} \end{matrix}\right.\)

If \(a=0\) then the system is equivalent to \(\left\{\begin{matrix} y=x-1\\ 0x=1 \end{matrix}\right.\). Since the equation \(0x=1\) is not strict, the system has no solution.

**Verse 2: **Know that the polynomial \(P(x)\) is divisible by \(xa\) if and only if \(P(a)=0\) (Bezout theorem). Find the values a, b such that the following polynomial is also divisible by \(x-1\) and \(x-2\):

\(P(x)=ax^4+(a-1)x^3+bx^2+3x+1\)

**Solution guide**

From the hypothesis we have \(\left\{\begin{matrix} P(1)=0\\ P(2)=0 \end{matrix}\right.<=>\left\{\begin{matrix} 2a+b=3\\ 24a+4b=1 \end{matrix}\right.\).

Solving this system by substitution method we get \(\left\{\begin{matrix} a=\frac{13}{16}\\ b=\frac{-37}{8} \end{matrix}\right .\)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Solve the following systems of equations using the substitution method:

a) \(\left\{\begin{matrix} x – y =3 & & \\ 3x-4y=2 & & \end{matrix}\right.\)

b) \(\left\{\begin{matrix} 7x – 3y =5 & & \\ 4x+y=2 & & \end{matrix}\right.\)

c) \(\left\{\begin{matrix} x +3y =-2 & & \\ 5x-4y=11 & & \end{matrix}\right.\)

**Verse 2:** Solve the following systems of equations using the substitution method:

a) \(\left\{\begin{matrix} 3x – 2y = 11 && \\ 4x – 5y = 3& & \end{matrix}\right.\)

b) \(\left\{\begin{matrix} \frac{x}{2}- \frac{y}{3} = 1& & \\ 5x – 8y = 3& & \end{matrix}\right.\ )

**Question 3: **Solve the system of equations by the substitution method:

a) \(\left\{\begin{matrix} x + y\sqrt{5} = 0& & \\ x\sqrt{5} + 3y = 1 – \sqrt{5}& & \end{matrix}\ right.\)

b) \(\left\{\begin{matrix} (2 – \sqrt{3})x – 3y = 2 + 5\sqrt{3}& & \\ 4x + y = 4 -2\sqrt{3} & & \end{matrix}\right.\)

**Question 4: **Solve the following systems of equations using the substitution method.

a) \(\left\{\begin{matrix} 3x – y = 5 & & \\ 5x + 2y = 23 & & \end{matrix}\right.\)

b) \(\left\{\begin{matrix} 3x +5y = 1 & & \\ 2x -y =-8 & & \end{matrix}\right.\)

c) \(\left\{\begin{matrix} \frac{x}{y} = \frac{2}{3}& & \\ x + y – 10 = 0 & & \end{matrix}\right .\)

### 3.2. Multiple choice exercises

**Question 1: **Find the solution of the following system of equations \(\left\{\begin{matrix} x+y=3\\ 2x-3y=1 \end{matrix}\right.\).

A. (1;1)

B. (1;0)

C. (1;2)

D. (2;1)

**Verse 2: **Find the values of a so that the following two systems of equations are equivalent:

\(\left\{\begin{matrix} x+3y=5\\ 2x-3y=1 \end{matrix}\right (I)\) and \(\left\{\begin{matrix} ax+ y=1\\ x+y=3 \end{matrix}\right.(II)\)

A. 0

B. 1

C. 2

D. 3

**Question 3: **Find the solution of the following system of equations \(\left\{\begin{matrix} x+y=2\\ 2x-y=1 \end{matrix}\right.\).

A. (-1;0)

B. (0;1)

C. (1;0)

D. (1;1)

**Question 4:** What is the number of solutions to the following system of equations \(\left\{\begin{matrix} 2x-y=1\\ -4x+2y=-2 \end{matrix}\right.\) ?

A. countless

B. 1

C. 2

D. inexperienced

**Question 5: **Find the integer values of m such that the following system of equations has a unique solution and that the solutions are positive numbers

\(\left\{\begin{matrix} xy=3\\ mx+y=3 \end{matrix}\right.\)

A. -1

B. 1

C. 0

D. 2

## 4. Conclusion

Through this lesson, you will learn a few things:

- State the substitution rule, determine the steps to solve the system of equations by the substitution method.
- Apply knowledge to solve some systems of equations by the substitution method.
- Know how to solve a system of equations using the substitution method.

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