Lecture 4 : System of Simultaneous Linear Equations

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Lecture 4
System of Simultaneous Linear Equations

1. In general, the system of two equations of 2 variables can be expressed in the form

$$
\left\{\begin{array}{l}
a_1 x+b_1 y=c_1, \\
a_2 x+b_2 y=c_2 .
\end{array}\right.
$$
2. To eliminating one variable for solving the system, we use (i) operations on equations as usual; (ii) substitution method. In many cases the method (i) is effective.
3. When $\frac{a-1}{a_2} \neq \frac{b_1}{b_2}$, the system has unique solution
$$
x=\frac{c_1 b_2-c_2 b_1}{a_1 b_2-a_2 b_1}, \quad y=\frac{a_1 c_2-a_2 c_1}{a_1 b_2-a_2 b_1} .
$$
4. When $\frac{a-1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ the system has two same equations, so it has infinitely many solutions, when $\frac{a-1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the two equations are inconsistent, so it has no solution.

Examples

Example 1. Solve the system of equations
$$
\left\{\begin{aligned}
\frac{x-y}{5}-\frac{x+y}{4} & =\frac{1}{2} \\
2(x-y)-3(x+y)+1 & =0 .
\end{aligned}\right.
$$

Solution (I) By operations on equations to eliminate a variable.
Simplifying the first equation, we have $4(x-y)-5(x+y)=10$, i.e.
$$x+9 y=-10 \mapsto ...(4.1)$$

Simplifying the second equation, we have
$$
x+5 y=1 \mapsto....(4.2)
$$

By $(4.1)-(4.2)$,
$$
4 y=-11, \therefore y=-\frac{11}{4} \text {. }
$$

From (4.2), $x=1-5 y=1+\frac{55}{4}=\frac{59}{4}$. Thus, $x=\frac{59}{4}, y=-\frac{11}{4}$.
(II) By substitution to eliminate a variable.
From the first equation we have
$$
x=-10-9 y .\mapsto [4.3]
$$

Substituting (4.3) into the second equation, we obtain
$$
\begin{gathered}
2(-10-9 y-y)-3(-10-9 y+y)+1=0, \\
4 y=-11, \therefore y=-\frac{11}{4} .
\end{gathered}
$$

By substituting it back into (4.3), we obtain $x=-10+\frac{99}{4}=\frac{59}{4}$. Thus,
$$
x=\frac{59}{4}, \quad y=-\frac{11}{4} .
$$

Example 2. Solve the system of equations
$$
\begin{aligned}
& 5.4 x+4.6 y=104, \\
& 4.6 x+5.4 y=96 .
\end{aligned}
$$

Solution Notice the feature of coefficients, by $(4.4)+(4.5)$, we obtain $10 x+$ $10 y=200$, therefore
$$
x+y=20 .
$$

By (4.4) - (4.5), it follows that $0.8 x-0.8 y=8$, therefore
$$
x-y=10 .
$$

By $\frac{1}{2}((4.6)+(4.7))$ and $\frac{1}{2}((4.6)-(4.7))$ respectively, we obtain
$$
x=15, \quad y=5 .
$$

Example 3. Solve the system of equations
$$
\left\{\begin{aligned}
x+2(5 x+y) & =16 \\
5 x+y & =7 .
\end{aligned}\right.
$$

Solution By using 7 to substitute $5 x+7$ in the first equation, we obtain $x+14=16$, therefore $x=2$.
Then from the second equation, $y=7-5 x=-3$.
Note: The example indicates that not only a variable but an expression can be substituted also.

Example 4. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{x}{2}=\frac{y}{3}=\frac{z}{5}, \\
x+3 y+6 z=15 .
\end{array}\right.
$$

Solution Let $t=\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$, then
$$
x=2 t, y=3 t, z=5 t .
$$

Substituting (4.8) into the first equation, we have $2 t+9 t+30 t=15$, i.e. $t=\frac{15}{41}$. Thus
$$
x=\frac{30}{41}, \quad y=\frac{45}{41}, \quad z=\frac{75}{41} .
$$

Example 5. Solve the system of equations
$$
\left\{\begin{array}{l}
x+y=5 \\
y+z=6 \\
z+x=7
\end{array}\right.
$$

Solution Let the given equations be labeled as
$$
\begin{aligned}
& x+y=5 \\
& y+z=6 \\
& z+x=7
\end{aligned}
$$

By $\frac{1}{2}((4.9)+(4.10)+(4.11))$, it follows that
$$
x+y+z=9 .
$$

Then (4.12) $-(4.9)$ yields $z=4$;
(4.12) - (4.10) yields $x=3$;
(4.12) - (4.11) yields $y=2$.

Example 6. Solve the system of the equations
$$
\left\{\begin{array}{l}
x+2 y=5, \\
y+2 z=8, \\
z+2 u=11, \\
u+2 x=6 .
\end{array}\right.
$$

Solution From the given equations we have the cyclic substitutions
$$
x=5-2 y, \quad y=8-2 z, \quad z=11-2 u, \quad u=6-2 x .
$$

By substituting them sequentially, we have
$$
\begin{aligned}
x & =5-2 y=5-2(8-2 z)=-11+4 z=-11+4(11-2 u)=33-8 u \\
& =33-8(6-2 x)=-15+16 x,
\end{aligned}
$$
therefore $x=16 x-15$, i.e. $x=1$, and then $u=4, z=3, y=2$.
Example 7. Solve the system of equations
$$
\begin{aligned}
& 5 x-y+3 z=a, \\
& 5 y-z+3 x=b, \\
& 5 z-x+3 y=c .
\end{aligned}
$$

Solution By $2 \times(4.13)+(4.14)-(4.15)$, it follows that
$$
14 x=2 a+b-c, \quad \therefore x=\frac{2 a+b-c}{14} .
$$

By $2 \times(4.14)+(4.15)-(4.13)$, it follows that
$$
14 y=2 b+c-a, \quad \therefore y=\frac{2 b+c-a}{14} .
$$

Similarly, by $2 \times(4.15)+(4.13)-(4.14)$, we have
$$
14 z=2 c+a-b, \therefore z=\frac{2 c+a-b}{14} .
$$

Example 8. Given that $x, y, z$ satisfy the system of equations
$$
\begin{aligned}
2000(x-y)+2001(y-z)+2002(z-x) & =0, \\
2000^2(x-y)+2001^2(y-z)+2002^2(z-x) & =2001,
\end{aligned}
$$
find the value of $z-y$.

Solution Let $u=x-y, v=y-z, w=z-x$. Then $u, v, w$ satisfy the following system of equations
$$
\begin{aligned}
u+v+w & =0, \\
2000 u+2001 v+2002 w & =0, \\
2000^2 u+2001^2 v+2002^2 w & =2001 .
\end{aligned}
$$

By $2001 \times(4.18)-(4.19)$, we obtain
$$
u-w=0, \text { i.e. } u=w .
$$

From (4.18) again, we have $v=-2 w$. By substituting it into (4.20), we have
$$
\begin{gathered}
\left(2000^2-2 \cdot 2001^2+2002^2\right) w=2001, \\
{[(2002+2001)-(2001+2000)] w=2001,} \\
2 w=2001, \quad \therefore z-y=-v=2 w=2001 .
\end{gathered}
$$

Example 9. Solve the system of equations for $(x, y)$, and find the value of $k$.
$$
\begin{aligned}
x+(1+k) y & =0, \\
(1-k) x+k y & =1+k, \\
(1+k) x+(12-k) y & =-(1+k) .
\end{aligned}
$$

Solution To eliminate $k$ from the equation, by $(4.22)+(4.23)$, we obtain
$$
2 x+12 y=0 \text {, i.e. } x=-6 y .
$$

By substituting (4.24) into (4.21), we have $(k-5) y=0$. If $k \neq 5$, then $y=0$ and so $x=0$ also. From (4.22) we have $k=-1$.
$$
\text { If } k=5,(4.22) \text { yields }(-4)(-6 y)+5 y=6 \text {, so } y=\frac{6}{29}, x=-\frac{36}{29} \text {. }
$$

$\boxed{  Testing ~~Questions  }$
(A)
1. (CHINA/1997) Given that $x=2, y=1$ is the solution of system
$$
\left\{\begin{array}{l}
a x+b y=7, \\
b x+c y=5,
\end{array}\right.
$$
then the relation between $a$ and $c$ is
(A) $4 \mathrm{a}+\mathrm{c}=9$;
(B) $2 \mathrm{a}+\mathrm{c}=9 ;$
(C) $4 a-c=9$;
(D) $2 a-c=9$.



2. If the system in $x$ and $y$
$$
\left\{\begin{array}{l}
3 x-y=5 \\
2 x+y-z=0 \\
4 a x+5 b y-z=-22
\end{array}\right.
$$
and the system in $x$ and $y$
$$
\left\{\begin{array}{l}
a x-b y+z=8 \\
x+y+5=c \\
2 x+3 y=-4
\end{array}\right.
$$
have a same solution, then $(a, b, c)$ is
(A) $(2,3,4)$;
(B) $(3,4,5)$;
(C) $(-2,-3,-4)$;
(D) $(-3,-4,-5)$.
3. Determine the values of $k$ such that the system of equations
$$
\left\{\begin{array}{l}
k x-y=-\frac{1}{3} \\
3 y=1-6 x
\end{array}\right.
$$
has unique solution, no solution, and infinitely many solutions respectively.
4. Given $\frac{a b}{a+b}=2, \frac{a c}{a+c}=5, \frac{b c}{b+c}=4$, find the value of $a+b+c$.
5. Solve the system of equations
$$
\left\{\begin{array}{l}
x-y-z=5 \\
y-z-x=1 \\
z-x-y=-15 .
\end{array}\right.
$$
6. Solve the system of equations
$$
\left\{\begin{array}{l}
x-y+z=1 \\
y-z+u=2 \\
z-u+v=3 \\
u-v+x=4 \\
v-x+y=5
\end{array}\right.
$$
7. Given
$$
\begin{aligned}
& \frac{1}{x}+\frac{2}{y}+\frac{3}{z}=0, \\
& \frac{1}{x}-\frac{6}{y}-\frac{5}{z}=0 .
\end{aligned}
$$

Find the value of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$.

8. (CHINA/2001) Given that the system of equations
$$
\left\{\begin{aligned}
m x+2 y & =10 \\
3 x-2 y & =0,
\end{aligned}\right.
$$
has integer solution, i.e. $x, y$ are both integers. Find the value of $m^2$.
9. As shown in the given figure, $a, b, c, d, e, f$ are all rational numbers, such that the sums of three numbers on each row, each column and each diagonal are equal. Find the value of $a+b+c+d+e+f$.

$$
\begin{array}{|c|c|c|}
\hline a& b & 6 \\
\hline c & d & e \\
\hline f & 7 & 2 \\
\hline
\end{array}
$$

10. Solve the system
$$
\begin{aligned}
x+y+z+u & =10, \\
2 x+y+4 z+3 u & =29, \\
3 x+2 y+z+4 u & =27, \\
4 x+3 y+z+2 u & =22 .
\end{aligned}
$$


Testing Questions
(B)
1. Given that the system of equations $\left\{\begin{array}{l}3 x+m y=7 \\ 2 x+n y=4\end{array}\right.$ has no solution, where $m, n$ are integers between -10 and 10 inclusive, find the values of $m$ and $n$.
2. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2} \\
\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3} \\
\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}
\end{array}\right.
$$


3. Solve the system
$$
\left\{\begin{array}{l}
x(y+z-x)=60-2 x^2, \\
y(z+x-y)=75-2 y^2, \\
z(x+y-z)=90-2 z^2 .
\end{array}\right.
$$
4. Find the values of $a$ such that the system of equations in $x$ and $y$
$$
\begin{array}{r}
x+2 y=a+6 \\
2 x-y=25-2 a
\end{array}
$$
has a positive integer solution $(x, y)$.
5. Solve the system of equations
$$
\begin{aligned}
& 2 x+y+z+u+v=16, \\
& x+2 y+z+u+v=17, \\
& x+y+2 z+u+v=19, \\
& x+y+z+2 u+v=21, \\
& x+y+z+u+2 v=23 .
\end{aligned}
$$