Solving Equations Reducible to Linear Form
Examples of Problem Solving
As a result of equivalent transformations, many equations are reduced to the linear form.
Example 1. Solve the equation (2x+3)(3x-2)-5x = 6x(x-1)
Solution:
$ (2x+3)(3x-2)-5x=6x(x-1) \iff 2x(3x-2)+3(3x-2)-5x = 6x^2-6x \iff $
$ \iff 6x^2-4x+6x-6-5x = 6x^2-6x \iff 6x^2-3x-6x^2+6x = 6 \iff $
$ \iff 3x = 6 \iff x = 2 $
Answer: x = 2
Example 2. Solve the equation $x+ \frac{1}{3}x = 11- \frac{1}{2}x $
Solution:
$ x+ \frac{1}{3}x = 11- \frac{1}{2}x \iff x+\frac{1}{3}x+ \frac{1}{2}x = 11 \iff (1+ \frac{1}{3} + \frac{1}{2})x = 11 \iff $
$ \iff \frac{11}{6}x = 11 \iff x = 11 \cdot \frac{6}{11} = 6 $
Answer: x = 6
Example 3. Solve the equation from the papyrus of Ahmes $ x+ \frac{2}{3}x+ \frac{1}{2}x+ \frac{1}{7}x=37 $
(ancient Egypt, 1700 BC)
Solution:
$ x+ \frac{2}{3}x+ \frac{1}{2}x+\frac{1}{7}x = 37 | \times 42 \iff 42x+2 \cdot 14x+21x+6x = 37 \cdot 42 \iff $
$ \iff 97x=1554 \iff x= \frac{1554}{97} = 16 \frac{2}{97} $
Answer: $ x = 16 \frac{2}{97}$
Example 4. Solve the equation $ \frac{3x-7}{3}-\frac{5x-11}{5} = 0 $
Solution:
$ \frac{3x-7}{3} - \frac{5x-11}{5} = 0 |× 15 \iff 5(3x-7)-3(5x-11)=0 \iff $
$ \iff 15x-35-15x+33 = 0 \iff 0x = 2 \iff x \in \Bbb \varnothing $
No roots.
Answer: $ x \in \Bbb \varnothing$
Example 5. Solve the equation $ \frac{2x+16}{2}-\frac{3x-7}{3} = 10 \frac{1}{3} $
Solution:
$$ \frac{2x+16}{2}-\frac{3x-7}{3} = 10 \frac{1}{3} | \times 6 \iff 3(2x+16)-2(3x-7) = 31 \cdot 2 \iff $$
$$ \iff 6x+48-6x+14 = 62 \iff 0x = 0 \iff x \in \Bbb R - any $$
Answer: $ x \in \Bbb R - any $
Example 6. Solve the equation |x+5| = 16
Solution:
$ |x+5| = 16 \iff \left[ \begin{array}{cc} x+5 = -16 \\ x+5 = 16 \end{array} \right. \iff \left[ \begin{array}{cc} x_1 = -21 \\ x_2=11 \end{array} \right. $
Answer: $ x_1 = -21, x_2 = 11 $
Example 7. Solve the equation |0,8x-4| = 0
Solution:
$ |0,8x-4| = 0 \iff 0,8x-4 = 0 \iff 0,8x = 4 \iff x = 4:0,8 = 5 $
Answer: x = 5
Example 8. Solve the equation |0,5x+2| = -3
Solution:
The absolute value cannot be negative.
$ |0,5x+2| = -3 \iff x \in \Bbb \varnothing $ - no roots
Answer: $ x \in \Bbb \varnothing $ - no roots
Example 9*. Solve the equation |5-|x+1||-3 = 2
Solution:
$ |5-|x+1||-3 = 2 \iff |5-|x+1|| = 5 \iff \left[ \begin{array}{cc} 5-|x+1| = -5 \\ 5-|x+1| = 5 \end{array} \right. \iff $
$ \iff \left[ \begin{array}{cc} |x+1| = 10 \\ |x+1|=0 \end{array} \right. \iff \left[ \begin{array}{cc} x+1 = -10 \\ x+1 = 10 \\ x+1 = 0 \end{array} \right. \iff \left[ \begin{array}{cc} x_1 = -11 \\ x_2=9 \\ x_3 = -1 \end{array} \right. $
Answer: $ x_1 = -11, x_2 = 9, x_3 = -1$
Example 10. Find the values of the parameter a such that x=-3 is the root of the equation a|3x+5|-6=2.
Solution:
Substitute the value of x into the equation:
$ a|3 \cdot (-3)+5|-6 = 2 \iff 4a-6 = 2 \iff 4a = 8 \iff a = 2 $
Answer: a = 2
