Page 51 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.5 Solve the equation x − 12x + 47x − 60x =0.
2
3
4
Solution:x − 12x + 47x − 60x =0 ⇔ x(x − 3x − 9x + 27x + 20x − 60) = 0 ⇔
2
2
3
4
3
2
3
4
2
2
3
2
x − 12x + 47x − 60x =0 ⇔ x(x − 3x − 9x + 27x + 20x − 60) = 0 ⇔
2
x[x (x − 3) − 9x(x − 3) + 20(x − 3)] = 0 ⇔ x(x − 3)(x − 4)(x − 5) = 0
x[x (x − 3) − 9x(x − 3) + 20(x − 3)] = 0 ⇔ x(x − 3)(x − 4)(x − 5) = 0 , which leads to
2
four solutions: x =0 or 3 or 4 or 5.
2.6 Given |x − 2| < 3, solve the equation |x +1| + |x − 3| + |x − 5| =8.
Solution 1: |x − 2| < 3 ⇒−1 <x< 5.
Then |x +1| + |x − 3| + |x − 5| =8 ⇒ x + 1+ |x − 3|− x +5 = 8 ⇒|x − 3| =2.
When x ≥ 3, x − 3= 2 ⇒ x =5 which does not satisfy the given inequality;
When x< 3, −(x − 3) = 2 ⇒ x =1 which satisfies the given inequality.
Solution 2: |x − 2| < 3 ⇒−1 <x< 5.
When −1 < x< 3, |x +1| + |x − 3| + |x − 5| =8 ⇒ (x + 1) − (x − 3) − (x − 5) = 8 ⇒ x =1;
When 3 ≤ x< 5, |x +1| + |x − 3| + |x − 5| =8 ⇒ (x +1)+(x − 3) − (x − 5) = 8 ⇒ x =5,
a contradiction to 3 ≤ x< 5, or x =5 does not satisfy the given inequality.

2.7 Solve the equation x|x|− 3|x|− 4= 0.

Solution: When x ≥ 0, x|x|− 3|x|− 4= 0 ⇒ x − 3x − 4= 0 ⇒ (x + 1)(x − 4) = 0 ⇒ x = −1
2
(deleted since x ≥ 0) or x =4.

When x< 0, x|x|− 3|x|− 4= 0 ⇒−x +3x − 4= 0 ⇒ x − 3x +4 = 0 which has no
2
2
solution since Δ=9 − 16 < 0.

As a conclusion, the original equation has a unique solution x =4.

2.8 We know that the equation system

3x + my − 5=0
x + ny − 4=0
has no solution, and m, n are integers whose absolute values less than 7, find the values
of m, n.

Download free eBooks at bookboon.com
51