Solving linear equations: Section 2
If equal terms appear on both sides of an equation,
then we may "cancel" them.
x + b + d = c + d.
d appears on both sides. Therefore, we may cancel them.
x + b = c.
Theoretically, we can say that we subtracted d from both sides.
Finally, on solving for x:
x = c − b.
Problem 17. Solve for x :
x² + x − 5 = x² − 3
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Do the problem yourself first!
Cancel the x²'s:
| x − 5 | = | −3 |
| x | = | −3 + 5 = 2. |
Problem 18. Solve for x :
x − a + b = a + b + c
Cancel the b's but not the a's. On the left is −a, but on the right is +a. They are not equal.
Transpose −a:
| x | = | a + a + c |
| x | = | 2a + c. |
The unknown on both sides
Example 3. Solve for x :
| 4x − 3 | = | 2x − 11. |
| 1. Transpose the x's to the left and the numbers to the right: | ||
| 4x − 2x | = | −11 + 3. |
| 2. Collect like terms, and solve: | ||
| 2x | = | −8 |
| x | = | −4. |
This is another example of doing algebra with your eyes. In the first line, you should see that 2x goes to the left as −2x, and that −3 goes to the right as +3.
As a general rule for solving any linear equation, we can now state the following:
Transpose all the terms that involve the unknown to the left, and add them;
transpose the remaining terms to the right;
make 1 the final coefficient of the unknown, by dividing or multiplying.
Problem 19. Solve for x :
| 15 + x | = | 7 + 5x |
| x − 5x | = | 7 − 15 |
| −4x | = | −8 |
| x | = | 2. |
Problem 20. Solve for x :
| 1.25x − 6 | = | x | ||
| 1.25x −x | = | 6 | ||
| (1.25 − 1)x | = | 6 | On combining the like terms on the left. |
|
| .25 x | = | 6 | ||
| x | = | 24. | On multiplying both sides by 4. |
|
Problem 21. Remove parentheses, add like terms, and solve for x :
| (8x − 2) + (3 − 5x) | = | (2x − 1) − (x − 3) |
| 8x − 2 + 3 − 5x | = | 2x − 1 − x + 3 |
| on removing the parentheses | ||
| 3x + 1 | = | x + 2 |
| 3x − x | = | 2 − 1 |
| 2x | = | 1 |
| x | = | 1 2 |
Simple fractional equations
| Example 4. | x 2 |
= 4. |
Since 2 divides on the left, it will multiply on the right:
| x | = | 2· 4 |
| = | 8. | |
Problem 22. Solve for x :
| −x 5 |
= | 3 |
| −x | = | 15 |
| x | = | −15. |
| Problem 23. | x 4 |
= | 1 2 |
|||
| x | = | 4· | 1 2 |
= 2. | ||
Example 5. Solve for x:
| 4 x |
= 5. |
Solution. In the standard form of a simple fractional equation, x is in the numerator. But we can easiy make that standard form by taking the reciprocal of both sides.
| x 4 |
= | 1 5 |
||
| This implies | ||||
| x | = | 4 · | 1 5 |
|
| = | 4 5 |
. |
| Problem 24. | 2 x |
= | 1 3 |
|
| x 2 |
= | 3 | ||
| x | = | 6. | ||
Example 6. Fractional coefficient.
| 3x 4 |
= | y |
| Since 4 divides on the left, it will multiply on the right: | ||
| 3x | = | 4y. |
| And since 3 multiplies on the left, it will divide on the right: | ||
| x | = | 4y 3 |
| In other words, | 3 4 |
goes to the other side as its reciprocal, | 4 3 |
. |
| Note that | 3 4 |
is the coefficient of x : |
| 3x 4 |
= | 3 4 |
x | . |
Coefficients go to the other side as their reciprocals!
| Problem 25. | 2x 3 |
= | a | |
| x | = | 3a 2 |
||
| Problem 26. | 5 8 |
x | = | a − b | ||
| x | = | 8 5 |
(a − b) | |||
| Problem 27. | 4 5 |
x + | 6 | = | 14 | ||
| 4 5 |
x | = | 14 − 6 = 8 | ||||
| x | = | 5 4 |
· 8 = 10. | ||||
| Problem 28. | A | = | ½xB | |
| Exchange sides: | ||||
| ½xB | = | A | ||
| x | = | 2A B |
||
The reciprocal of ½ is 2.
Problem 29. The Celsius temperature C is related to the Fahrenheit temperature F by this formula,
| F | = | 9 5 |
C + 32 |
a) What is the Fahrenheit temperature when the Celsius temperature
a) is 10°?
| F | = | 9 5 |
· 10 + 32 |
| = | 18 + 32 | ||
| = | 50° | ||
b) Solve the formula for C.
| 9 5 |
C + | 32 | = | F | |
| 9 5 |
C | = | F − 32 | ||
| C | = | 5 9 |
(F − 32) | ||
c) What is the Celsius temperature when the Fahrenheit temperature
c) is 68°?
20°. This is Problem 13 of Lesson 1.
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