Linear Equations in One Variable: Definitions, Methods, and Applications
Linear equations in one variable form the foundation of algebra and are essential for solving real-world mathematical problems. A linear equation in one variable is an algebraic statement that equates two expressions containing only one unknown variable, typically represented as x, y, or z. These equations involve no exponents, square roots, or complex operations—just straightforward arithmetic relationships that can be manipulated to find the value of the variable.
Linear Equations
A statement of equality of two algebraic expressions in or more variables is called a linear equation. These are equations with just a plain old variable like "x", rather than something more complicated like x² or x/y or square roots or such.
Example 1: Solve x + 6 = –3
Solution:
Here, we want to get the x by itself; that is, we want to get "x" on one side of the "equals" sign, and some number on the other side. Since we want just x on the one side, this means that we don't like the "plus six" that's currently on the same side as the x. Since the 6 is added to the x, we need to get rid of it. That is, I will need to subtract a 6 from the x in order to "undo" having added a 6 to it.
Whatever you do to an equation, do the SAME thing to BOTH sides of that equation!
CONSTANTS AND VARIABLES
What is a CONSTANT?
A CONSTANT is a quantity whose value remains the same throughout a particular problem.
What is a VARIABLE?
A VARIABLE is a quantity whose value is free to vary.
Examples:
1. x + 1 = 2 and
2. 2y + 3 = 5
In 1 above, x + 1 is an algebraic expression in the variable x. We read it as "x plus 1 is equal to 2". x is some number, called variable.
In 2 above, 2y + 3 is an algebraic expression in the variable y, we read it as "2 times y or 2y plus 3 is equal to 5". y is some number, called variable.
Types of Constants
There are two kinds of constants - fixed and arbitrary:
FIXED Constants: Numbers such as 7, -3, 1/2, and π are examples of FIXED constants. Their values never change. In 5x + 7 = 0, the numbers 0, 5, and 7, are fixed constants.
ARBITRARY Constants: Arbitrary constants can be assigned different values for different problems. Arbitrary constants are indicated by letters - quite often letters at the beginning of the alphabet such as a, b, c, and d.
Understanding Variables
A variable may have one value or it may have many values in a discussion. The letters at the end of the alphabet, such as x, y, z are usually used to represent variables. In 5x + 7, the letter x is the variable.
If x = 1, then 5x + 7 = 5 + 7 = 12
If x = 2, then 5x + 7 = 5(2) + 7 = 10 + 7 = 17
and so on for as many values of x as we desire to select.
If the expression 5x + 7 is set equal to some particular number, say -23, then the resulting equality:
5x + 7 = -23
holds true for just one value of x. The value is -6, since 5(-6) + 7 = -23
Variable Terms and Constant Terms
In an algebraic expression, terms that contain a variable are called VARIABLE TERMS. Terms that do not contain a variable are CONSTANT TERMS.
The expression 5x + 7 contains one variable term and one constant term. The variable term is 5x, while 7 is the constant term. In ax + b, ax is the variable term and b is the constant term.
A variable term often is designated by naming the variable it contains. In 5x + 7, 5x is the x-term. In ax + by, ax is the x-term, while by is the y-term.
Solving Linear Equation
1. What is Solving a Linear Equation?
Solving an equation means to find a value of the variable which satisfies the equation.
2. How do we solve the equation?
Stepwise RULES FOR SOLVING AN EQUATION:
- Same quantity can be added to both sides of an equation without changing the equality;
- Same quantity can be subtracted from both sides of an equation without changing the equality;
- Both sides of an equation may be multiplied by a same non-zero number without changing the equality;
- Both sides of an equation may be divided by a same non-zero number without changing the equality.
The Three Components of an Equation
Every equation has a Left hand side, the equality sign '=' and the Right hand side. The three components in the equation x + 1 = 2 are:
| Linear Equation | L.H.S. | Equality | R.H.S. |
|---|---|---|---|
| x + 1 = 2 | x + 1 | = | 2 |
L.H.S = Left Hand Side, R.H.S = Right Hand Side
SOLUTION/ROOT of an Equations
The value of x, i.e. some number for x, which makes the equation a true statement is called solution or root of the equation.
In simple words, if the L.H.S. and R.H.S become equal for some number plugged in for x, then the number, also called value, is the solution or root of the equation.
In x + 1 = 2, what should be plugged in for x so that L.H.S. becomes equal to R.H.S?
It is 1, i.e. if 1 is plugged in for x, the two sides become equal.
This number or value 1 for x is called root or solution of the equation.
Transposition Rule
The transposition rule applies on addition and subtraction.
Transposition Rule: Terms can be transposed (shifted) between either sides of the equality symbol "=" with a change in sign of the transposed terms.
e.g. x + 1 = 2.
1 can be transposed to right side by inversing its sign i.e. x = -1 + 2 so that we have x = 1
Transposition conforms with the rule: "same numbers can be added on both sides of an equation"
So, in x + 1 = 2, to find x, we need to get rid of 1 on the left hand side, to do this add -1 on both sides of the equation. x + 1 - 1 = -1 + 2, x = 1
Operations on Linear Equestions
Addition
Find the value of x in the equation: x - 3 = 12
As in any equation, we must isolate the variable on either the right or left side. In this problem, we leave the variable on the left and perform the following steps:
1. Add 3 to both members of the equation, as follows:
x - 3 + 3 = 12 + 3
In effect, we are "undoing" the subtraction indicated by the expression x - 3, for the purpose of isolating x in the left member.
2. Combining terms, we have
x = 15
Subtraction
Find the value of x in the equation: x + 14 = 24
1. Subtract 14 from each member. In effect, this undoes the addition indicated in the expression x + 14.
x + 14 - 14 = 24 - 14
2. Combining terms, we have
x = 10
Multiplication
Find the value of y in the equation: y/5 = 10
1. The only way to remove the 5 so that the y can be isolated is to undo the indicated division. Thus we use the inverse of division, which is multiplication. Multiplying both members by 5, we have the following:
5(y/5) = 5(10)
2. Performing the indicated multiplications, we have
y = 50
Division
Find the value of x in the equation: 3x = 15
1. The multiplier 3 may be removed from the x by dividing the left member by 3. This must be balanced by dividing the right member by 3 also, as follows:
3x/3 = 15/3
2. Performing the indicated divisions, we have
x = 5
Solutions Requiring More Than One Operation
Most equations involve more steps in their solutions than the simple equations already described, but the basic operations remain unchanged. If the basic axioms are kept well in mind, these more complicated equations will not become too difficult. Equations may require one or all of the basic operations before a solution can be obtained.
Subtraction and Division:
Find the value of x in the following equation: 2x + 4 = 16
1. The term containing x is isolated on the left by subtracting 4 from the left member. This operation must be balanced by also subtracting 4 from the right member, as follows:
2x + 4 - 4 = 16 - 4
2. Performing the indicated operations, we have
2x = 12
3. The multiplier 2 is removed from the x by dividing both sides of the equation by 2, as follows:
2x/2 = 12/2
x = 6
Addition, Multiplication and Division
Find the value of y in the following equation: (3y/2) - 4 = 11
1. Isolate the term containing y on the left by adding 4 to both sides, as follows:
(3y/2) - 4 + 4 = 11 + 4
(3y/2) = 15
2. Since the 2 will not divide the 3 exactly, multiply the left member by 2 in order to eliminate the fraction. This operation must be balanced by multiplying the right member by 2, as follows:
2 × (3y/2) = 2 × 15
3y = 30
3. Divide both members by 3, in order to isolate the y in the left member, as follows:
3y/3 = 30/3
y = 10
Equestions Having The Variable in More Than One Term
Find the value of x in the following equation: (3x/4) + 2x = 12 - x
1. Rewrite the equation with no terms containing the variable in the right member. This requires adding x to the right member to eliminate the -x term, and balance requires that we also add x to the left member, as follows:
(3x/4) + 2x + x = 12 - x + x
(3x/4) + 2x = 12
2. Since the 4 will not divide the 3 exactly, it is necessary to multiply the first term by 4 to eliminate the fraction. However, notice that this multiplication cannot be performed on the first term only; any multiplier which is introduced for simplification purposes must be applied to the entire equation. Thus each term in the equation is multiplied by 4, as follows:
4 × (3x/4) + 4 × 2x = 4 × 12
3x + 8x = 48
3. Add the terms containing x and then divide both sides by 11 to isolate the x in the left member, as follows:
11x = 48
x = 48/11
Equations with Literal Coefficients
As stated earlier, the first letters of the alphabet usually represent known quantities (constants), and the last letters represent unknown quantities (variables). Thus, we usually solve for x, y, or z.
An equation such as ax - 8 = bx - 5 has letters as coefficients. Equations with literal coefficients are solved in the same way as equations with numerical coefficients, except that when an operation cannot actually be performed, it merely is indicated.
In solving for x in the equation: ax - 8 = bx - 5
Subtract bx from both members and add 8 to both members. The result is:
ax - bx = 8 - 5
Since the subtraction on the left side cannot actually be performed, it is indicated. The quantity, a - b, is the coefficient of x when terms are collected. The equation takes the form:
(a - b)x = 3
Now divide both sides of the equation by a - b. Again the result can be only indicated. The solution of the equation is:
x = 3/(a - b)
In solving for y in the equation: ay + b = 4
Subtract b from both members as follows:
ay = 4 - b
Dividing both members by a, the solution is:
y = (4 - b)/a
Removing Signs of Grouping
If signs of grouping appear in an equation they should be removed in the manner indicated. For example, solve the equation:
5 = 24 - [x - 12(x - 2) - 6(x - 2)]
Notice that the same expression, x - 2, occurs in both parentheses. By combining the terms containing (x - 2), the equation becomes:
5 = 24 - [x - 18(x - 2)]
Next, remove the parentheses and then the bracket, obtaining:
5 = 24 - [x - 18x + 36]
= 24 - [36 - 17x]
= 24 - 36 + 17x
= -12 + 17x
Subtracting 17x from both members and then subtracting 5 from both members, we have:
-17x = -12 - 5
-17x = -17
Divide both members by -17. The solution is:
x = 1
Equations Containing Fractions
To solve for x in an equation with fractions, first clear the equation of fractions. To do this, find the least common denominator of the fractions. Then multiply both sides of the equation by the LCD.
Example: Solve equation with fractions where LCD is 12
The least common denominator of 3, 12, 4, and 2 is 12. Multiply both sides of the equation by 12. The resulting equation is:
8x + x - 12 = -3 + 8x
Subtract 6x from both members, add 12 to both members, and collect like terms as follows:
9x - 6x = 12 + 3
3x = 15
The solution is:
x = 5
Practice Problems with Solutions
Problem 1: Solve 2x = 5
Since the x is multiplied by 2, I need to divide both sides by 2:
2x/2 = 5/2
Then the solution is x = 5/2 or x = 2.5
Problem 2: Solve x/5 = -6
Since the x is divided by 5, multiply both sides by 5:
x = -6 × 5
Then the solution is x = -30
Problem 3: Check x = -9 as a solution for x + 6 = -3
x + 6 = -3
[-9] + 6 ?=? -3
-9 + 6 ?=? -3
6 - 9 ?=? -3
-3 = -3 ✓
Problem 4: Check x = -2 for x - 3 = -5
x - 3 = -5
[-2] - 3 ?=? -5
-2 - 3 ?=? -5
-2 + (-3) ?=? -5
-5 = -5 ✓
Problem 5: Check x = 2.5 for 2x = 5
2x = 5
2[2.5] ?=? 5
5 = 5 ✓
Problem 6: Check x = -30 for x/5 = -6
x/5 = -6
[-30]/5 ?=? -6
-6 = -6 ✓
Problem 7: Solve (3/5)x = 10
Since x is multiplied by 3/5, I'll want to multiply both sides by 5/3, to cancel off the fraction on the x. Many students find it helpful to also turn the 10 into a fraction, by putting it over 1.
(5/3) × (3/5)x = (5/3) × (10/1)
x = 50/3
Then the solution is x = 50/3
Problem 8: Solve: 2x - 3 = 18
We have:
2x - 3 = 18
2x = 18 + 3
2x = 21
x = 21/2 = (9 × 3)/2 = 27/2
x = 27
Problem 9: Solve: (5x/3) - 1 = 6/5
(5x/3) - 1 = 6/5
(5x/3) = 1 + 6/5
(5x/3) = (5 + 6)/5 = 11/5
x = (11/5) × (3/5) = 33/25
Problem 10: Solve the equation: (3x/7) + 17 = 7
(3x/7) + 17 = 7
(3x + 17 × 7)/7 = 7
(3x + 119)/7 = 7
3x + 119 = 7 × 7 = 49
3x = 49 - 119 = -70
x = -70/3
x = 2
Problem 11: Solve the equation: y + 1.6 = 1.5
y + 1.6 = 1.5
or y = 1.5 - 1.6 = -0.1
y = -0.1 (or) -2.40
y = 2.4
Problem 12: Solve the equation: x + (1/3) = (7/15) - 1
x + (1/3) = (7/15) - 1
or x = (7/15) - 1 - (1/3) [subtracting 1 from both sides]
or x = (7 - 15)/15 - (1/3)
or x = -8/15 - 1/3
x = (-8 × 3)/(15 × 3) = -8/5
When Variables are Present on Both Sides of The Equation
Problem 1: Solve: 8x + 4 = 3(x - 1) + 7
8x + 4 = 3(x - 1) + 7
or, 8x + 4 = 3x - 3 + 7
or, 8x - 3x = -3 + 7 - 4
or, 5x = 0
x = 0
Check:
L.H.S = 8 × 0 + 4 = 4
R.H.S = 3(0 - 1) + 7 = -3 + 7 = 4
L.H.S = R.H.S ✓
Problem 2: Solve: x = 5(x - 10) + 4
x = 5(x - 10) + 4
or, x = 5x - 50 + 4
or, 5x - x = 50 - 4
or, 4x = 46
x = 40
Check:
L.H.S = 40
R.H.S = 5(40 - 10) + 4 = 5 × 30 + 4 = 150 - 50 = 40
L.H.S = R.H.S ✓
Problem 3: Solve: (2x/3) - (1/3) = (7x/15) + 3
(2x/3) - (1/3) = (7x/15) + 3
or, (2x/3) - (7x/15) = 3 + (1/3)
or, (10x - 7x)/15 = (9 + 1)/3
or, 3x/15 = 10/3
or, x = (10 × 15)/(3 × 3) = 150/9 = 50/3
x = 10
Check:
L.H.S = (2 × 10)/3 - 1/3 = 20/3 - 1/3 = (20 - 1)/3 = 19/3
R.H.S = (7 × 10)/15 + 3 = 70/15 + 3 = 14/3 + 9/3 = 23/3
L.H.S = R.H.S ✓
Reducing and Simplifying
The best method to solve linear equations is to reduce the equation to the simplest form and then solve for the value of the unknown variable. The 2 ways to simplify the equation are:
- Multiplying both sides of the equation by the LCM of the denominators of the terms in the expressions of the equation.
- Opening brackets and combining like terms on both sides of the equation.
Problem 1: Solve: x - (x/2) - (x/4) = (8x - 2x)/3 + (x - 2)/24 - 6/3
We have:
x - (x/2) - (x/4) = (8x - 2x)/3 + (x - 2)/24 - 6/3
x - (x/2) - (x/4) = (6x/3) + (x - 2)/24 - 2
(24x - 12x - 6x)/24 = (8 × 6x + x - 2 - 48)/24
The denominators on the two sides of this equation are 3, 4 and 24. Their L.C.M. is 24
Multiplying both sides of this equation by 24, we get:
24x - 24(x/2) - 24(x/4) = 24(6x/3) + 24(x - 2)/24 - 24(2)
24x - 12x - 6x = 48x + (x - 2) - 48
6x = 48x + x - 2 - 48
6x = 49x - 50
6x - 49x = -50
-43x = -50 [Transposing 7x to LHS and -64 to RHS]
x = -10
Thus, x = -10 is the solution of the given equation.
Problem 2: Solve: (x - 2)/3 + (x - 1)/5 = (3x - 1)/4 - 1
Multiplying both sides by 60 i.e. the LCM of 3, 5 and 4, we get:
20(x - 2) + 12(x - 1) = 15(3x - 1) - 60
20x - 40 + 12x - 12 = 45x - 15 - 60
32x - 52 = 45x - 75
32x - 45x = -75 + 52
-13x = -23
x = 23/13 [Dividing both sides by -13]
x = -133/7 = 19
Thus, x = 19 is the solution of the given equation.
Problem 3: Solve: (x/2) - (1/5) = (x/3) - (1/4)
(x/2) - (1/5) = (x/3) - (1/4)
or (x/2) - (x/3) = (1/5) - (1/4)
or (3x - 2x)/6 = (4 - 5)/20
or x/6 = -1/20
or x = (-1 × 6)/20 = -6/20 = -3/10
x = 2.7
Problem 4: Solve: (3t - 2)/4 - (2t + 3)/3 = (2/3) - t
(3t - 2)/4 - (2t + 3)/3 = (2/3) - t
or, (3(3t - 2) - 4(2t + 3))/12 = (2 - 3t)/3
or, 3(3t - 2) - 4(2t + 3) = 4(2 - 3t)
or, 9t - 6 - 8t - 12 = 8 - 12t
or, t - 18 = 8 - 12t
or, 13t = 26
or, t = 26/13 = 2
t = 2/13
Problem 5: Solve: m - (m - 1)/2 = 1 - (m - 2)/3
m - (m - 1)/2 = 1 - (m - 2)/3
or, (6m - 3(m - 1))/6 = (3 - (m - 2))/3
or, 6m - 3m + 3 = 6 - 2m + 4 [Opening the brackets]
or, 3m + 3 = 10 - 2m
or, 5m = 7
m = 7/5
Problem 6: Solve: 15y - 4(y - 6) = 2(y - 9) + 0
15y - 4(y - 6) = 2(y - 9) + 0
or, 15y - 4y + 24 = 2y - 18 + 0
or, 11y + 24 = 2y - 18
or, 11y - 2y = -18 - 24
or, 9y = -42
or, y = -42/9 = -14/3
y = -2/3
Reducing Equation to Linear Form
(ax + b)/(cx + d) = m/n
If (ax + b)/(cx + d) = m/n is an equation in variable x, then:
(ax + b)/(cx + d) = m/n ⟹ n(ax + b) = m(cx + d)
which is a linear equation.
The process of obtaining the above linear equation from (ax + b)/(cx + d) = m/n is called cross multiplication.
Problem 1: Solve the equation: (7y - 4)/(y + 2) = -4/3
(7y - 4)/(y + 2) = -4/3
or, 3(7y - 4) = -4(y + 2)
or, 21y - 12 = -4y - 8
or, 21y + 4y = 12 - 8
or, 25y = 4
or, y = 4/25 = -20/-25 = -4/5
y = -4/5
Problem 2: Solve: (4x - 5)(5x - 7) = (2x - 7)(x + 1)
We have:
(4x - 5)(5x - 7) = (2x - 7)(x + 1)
20x² - 28x - 25x + 35 = 2x² + 2x - 7x - 7
20x² - 53x + 35 = 2x² - 5x - 7
20x² - 2x² - 53x + 5x = -7 - 35 [By cross-multiplication]
18x² - 48x = -42
-4x = -6
x = -6/-4 = 3/2 [Dividing both sides by -4]
Hence, x = 3/2 is the solution of the given equation.
Problem 3: Solve: (3x - 5)/(2x - 7) = 4
We have:
(3x - 5)/(2x - 7) = 4/1
1(3x - 5) = 4(2x - 7) [By cross-multiplication]
3x - 5 = 8x - 28
3x - 8x = -28 + 5 [Using transposition]
-5x = -23
x = -23/-5 = 23/5 [Dividing both sides by -5]
Hence, x = -23/5 is the solution of the given equation.
Problem 4: Solve the equation: (8x/3) = (3x/2)
(8x/3) = (3x/2)
or, 8x × 2 = 3x × 3
or, 16x = 9x
or, 16x - 9x = 0
or, 7x = 0
or, x = 3/2
Problem 5: Solve the equation: z/(z - 15) = 4/9
z/(z - 15) = 4/9
or, 9z = 4(z - 15)
or, 9z = 4z - 60
or, 9z - 4z = -60
or, 5z = -60
or, z = -12
z = -12
Problem 6: Solve: (2/x) + (3/(3 - x)) + (5/(4 - x)) = x/5, where x ≠ 0, x ≠ 3 and x ≠ 4
(2/x) + (3/(3 - x)) + (5/(4 - x)) = x/5
(2(3 - x)(4 - x) + 3x(4 - x) + 5x(3 - x))/(x(3 - x)(4 - x)) = x/5
(2(12 - 7x + x²) + 3x(4 - x) + 5x(3 - x))/(x(12 - 7x + x²)) = x/5
(24 - 14x + 2x² + 12x - 3x² + 15x - 5x²)/(x(12 - 7x + x²)) = x/5
(24 - 14x + 13x + 2x²)/(x(12 - 7x)) = x/5
5(24 - x - 2x²) = x(12 - 7x)
120 - 5x - 10x² = 12x - 7x²
120 - 5x - 12x = 10x² - 7x²
120 - 17x = 3x²
3x² + 17x - 120 = 0
or x = 10/3
Simplifying Decimal Equations
Problem 7: Simplify and solve the linear equation: 0.25(4f - 3) = 0.05(10f - 9)
0.25(4f - 3) = 0.05(10f - 9)
(1/4)(4f - 3) = (1/20)(10f - 9)
Multiplying both sides by 20, we obtain:
5(4f - 3) = 10f - 9
20f - 15 = 10f - 9 [Opening the brackets]
20f - 10f = -9 + 15
10f = 6
f = 6/10 = 3/5 = 0.6
Problem 8: Simplify and solve the linear equation: 3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17
3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17
15z - 21 - 18z + 22 = 32z - 52 - 17 [Opening the brackets]
-3z + 1 = 32z - 69
-3z - 32z = -69 - 1
-35z = -70
z = -70/-35 = 2
Problem 9: Simplify and solve the linear equation: 15(y - 4) - 2(y - 9) + 5(y + 6) = 0
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
15y - 60 - 2y + 18 + 5y + 30 = 0 [Opening the brackets]
18y - 12 = 0
18y = 12
y = 12/18 = 2/3
Problem 10: Simplify and solve the linear equation: 3(t - 3) = 5(2t + 1)
3(t - 3) = 5(2t + 1)
3t - 9 = 10t + 5 [Opening the brackets]
-9 - 5 = 10t - 3t
-14 = 7t
t = -14/7 = -2
Problem 11: Simplify and solve the linear equation: m - (m - 1)/2 = 1 - (m - 2)/3
m - (m - 1)/2 = 1 - (m - 2)/3
L.C.M. of the denominators, 2 and 3, is 6.
Multiplying both sides by 6, we obtain:
6m - 3(m - 1) = 6 - 2(m - 2)
6m - 3m + 3 = 6 - 2m + 4 [Opening the brackets]
6m - 3m + 2m = 6 + 4 - 3
5m = 7
m = 7/5
Problem 12: Solve the linear equation: (3t - 2)/4 - (2t + 3)/3 = (2/3) - t
(3t - 2)/4 - (2t + 3)/3 = (2/3) - t
L.C.M. of the denominators, 3 and 4, is 12.
Multiplying both sides by 12, we obtain:
3(3t - 2) - 4(2t + 3) = 8 - 12t
9t - 6 - 8t - 12 = 8 - 12t [Opening the brackets]
9t - 8t + 12t = 8 + 6 + 12
13t = 26
t = 26/13 = 2
Problem 13: Solve the linear equation: (x - 5)/3 = (x - 3)/5
(x - 5)/3 = (x - 3)/5
L.C.M. of the denominators, 3 and 5, is 15.
Multiplying both sides by 15, we obtain:
5(x - 5) = 3(x - 3)
5x - 25 = 3x - 9 [Opening the brackets]
5x - 3x = 25 - 9
2x = 16
x = 8
Problem 14: Solve the linear equation: x + (7 - 8x)/3 = (17/6) - (5x/2)
x + (7 - 8x)/3 = (17/6) - (5x/2)
L.C.M. of the denominators, 2, 3, and 6, is 6.
Multiplying both sides by 6, we obtain:
6x + 2(7 - 8x) = 17 - 15x
6x + 14 - 16x = 17 - 15x
6x - 16x + 15x = 17 - 14
5x = 3
x = -25/-5 = 5
Problem 15: Solve the linear equation: (n/2) - (3n/4) + (5n/6) = 21
(n/2) - (3n/4) + (5n/6) = 21
L.C.M. of the denominators, 2, 4, and 6, is 12.
Multiplying both sides by 12, we obtain:
6n - 9n + 10n = 252
7n = 252
n = 252/7 = 36
Problem 16: Solve the linear equation: (x/2) - (1/5) = (x/3) + (1/4)
(x/2) - (1/5) = (x/3) + (1/4)
L.C.M. of the denominators, 2, 3, 4, and 5, is 60.
Multiplying both sides by 60, we obtain:
60(x/2) - 60(1/5) = 60(x/3) + 60(1/4)
30x - 12 = 20x + 15 [Opening the brackets]
30x - 20x = 15 + 12
10x = 27
x = 27/10