Back

System of Linear Equations

[1324]⋅[12][710][1324]⋅[xy][710]x+3y=72x+4y=10\begin{gather*} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \end{bmatrix} \begin{bmatrix} 7 \\ 10 \end{bmatrix} \\[6pt] \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} 7 \\ 10 \end{bmatrix} \\[6pt] x + 3y = 7 \\ 2x + 4y = 10 \end{gather*}

Properties of Linear Equations

1. Multiplying or Dividing an Equation

We can multiply or divide both sides of an equation by the same constant.
The solution will remain unchanged.

2(x+3y)=7(2)2x+6y=142(1)+6(2)=142+12=14\begin{gather*} 2(x + 3y) = 7(2) \\ 2x + 6y = 14 \\ 2(1) + 6(2) = 14 \\ 2 + 12 = 14 \end{gather*}
2x+4y=10×36x+12y=306(1)+12(2)=30\begin{gather*} 2x + 4y = 10 \qquad \times 3 \\ 6x + 12y = 30 \\ 6(1) + 12(2) = 30 \end{gather*}

2. Swapping the Order of Equations

We can change the order of equations (swap rows).
This does not change the solution.

x+3y=7(1)2x+4y=10(2)\begin{gather*} x + 3y = 7 \quad (1) \\ 2x + 4y = 10 \quad (2) \end{gather*} 2x+4y=10(1)x+3y=7(2)\begin{gather*} 2x + 4y = 10 \quad (1) \\ x + 3y = 7 \quad (2) \end{gather*}

3. Adding or Combining Equations

We can add two or more equations together.
The resulting equation will still have the same solution.

2x+6y=148x+16y=4010x+22y=5410(1)+22(2)=5410+44=54\begin{gather*} 2x + 6y = 14 \\ 8x + 16y = 40 \\ \hline 10x + 22y = 54 \\ 10(1) + 22(2) = 54 \\ 10 + 44 = 54 \end{gather*}

Practice

1.

[1379][xy][416]x+3y=47x+9y=16\begin{gather*} \begin{bmatrix} 1 & 3 \\ 7 & 9 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} 4 \\ 16 \end{bmatrix} \\ x + 3y = 4 \\ 7x + 9y = 16 \end{gather*}

We want to eliminate x from the equations.

Multiply the first equation by −7:

−7x−21y=−287x+9y=160x−12y=−12y=1\begin{gather*} -7x - 21y = -28 \\ 7x + 9y = 16 \\ \hline 0x - 12y = -12 \\ y = 1 \end{gather*}

Another method is multiplying the first equation by 3, then solving for y.

2.

[1379][x y][630]x+3y=67x+9y=30\begin{gather*} \begin{bmatrix} 1 & 3 \\ 7 & 9 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} \begin{bmatrix} 6 \\ 30 \end{bmatrix} \\ x + 3y = 6 \\ 7x + 9y = 30 \end{gather*}

Eliminate x:

−7x−21y=−427x+9y=300x−12y=−12y=1\begin{gather*} -7x - 21y = -42 \\ 7x + 9y = 30 \\ \hline 0x - 12y = -12 \\ y = 1 \end{gather*}

Substitute (y = 1) into one equation:

7x+9y=307x+9(1)=307x+9=307x=21x=3\begin{gather*} 7x + 9y = 30 \\ 7x + 9(1) = 30 \\ 7x + 9 = 30 \\ 7x = 21 \\ x = 3 \end{gather*}

3.

[1309][xy][718] \begin{gather*} \begin{bmatrix} 1 & 3 \\ 0 & 9 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} 7 \\ 18 \end{bmatrix} \end{gather*}

Convert to equations:

x+3y=79y=18 \begin{gather*} x + 3y = 7 \\ 9y = 18 \end{gather*}

Solve:

y=2 y = 2 x+3(2)=7 x + 3(2) = 7 x=1 x = 1

Solving Using Row Operations

Matrix Form

[1324][xy][710] \begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} \begin{bmatrix} 7\\ 10 \end{bmatrix}

Augmented Matrix

[1372410] \left[ \begin{array}{cc|c} 1 & 3 & 7\\ 2 & 4 & 10 \end{array} \right]

Row Operation

−2R1+R2→R2 -2R_1 + R_2 \to R_2

After Elimination

[1370−2−4] \left[ \begin{array}{cc|c} 1 & 3 & 7\\ 0 & -2 & -4 \end{array} \right]

Back Substitution

−2y=−4 -2y = -4 y=2 y = 2 x+3(2)=7 x + 3(2) = 7 x=1 x = 1

Final Answer

(x,y)=(1,2) (x,y) = (1,2)

Solving a System of Three Equations

[123502−214][xyz][141112] \begin{bmatrix} 1 & 2 & 3 \\ 5 & 0 & 2 \\ -2 & 1 & 4 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} \begin{bmatrix} 14\\ 11\\ 12 \end{bmatrix}

Augmented Matrix

[1231450211−21412] \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 14\\ 5 & 0 & 2 & 11\\ -2 & 1 & 4 & 12 \end{array} \right]

Row Operations

R2−5R1→R2 R_2 - 5R_1 \to R_2 R3+2R1→R3 R_3 + 2R_1 \to R_3

After Elimination

[123140−10−13−59051040] \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 14\\ 0 & -10 & -13 & -59\\ 0 & 5 & 10 & 40 \end{array} \right]

Continue Elimination

−110R2→R2 -\frac{1}{10}R_2 \to R_2 R3−5R2→R3 R_3 - 5R_2 \to R_3

Upper Triangular Matrix

[1231401131059100072212] \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 14\\ 0 & 1 & \frac{13}{10} & \frac{59}{10}\\ 0 & 0 & \frac{7}{2} & \frac{21}{2} \end{array} \right]

Back Substitution

72z=212 \frac{7}{2}z=\frac{21}{2} z=3 z=3 y+1310(3)=5910 y+\frac{13}{10}(3)=\frac{59}{10} y=2 y=2 x+2(2)+3(3)=14 x+2(2)+3(3)=14 x=1 x=1

Final Solution

(x,y,z)=(1,2,3) (x,y,z) = (1,2,3)