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LINEAR EQUATIONS
Definitions: An equality between two algebraic expressions, which is satisfied only by
certain values of its variables, is known as an algebraic equation. If the equality is true for
any value of the variables then it is called an identity. We will refer to the variables of the
equation as unknowns, and to the values satisfying the equality as solutions or roots. Solving
an equation means to find all its solutions. Two equations are said to be equivalent if they
have the same solutions. If P is a polynomial, the equation P=0 is called a polynomial
equation. The degree of P is the degree or order of the equation. In particular, if P=ax+b,
then the equation ax+b=0 is called a linear equation. Examples:
x3 y  y 2  y  3 is a polynomial equation of degree 4 in two variables.
2( x  1)  2 x  2 is an identity.
7
3
x   0 is a linear equation in one variable.
2
5
x 2  3  1  x is an irrational equation.
The equations 3x-2=1 and 4x-3=1 are equivalent. Notice that x=1 is the only solution for
both equations.
Addition Principle of Equality: If we add (or subtract) the same number or expression to
both sides of an equation we get an equivalent equation.
Example: The equation x-7=3 is equivalent to x-7+7=3+7, that is x=10, which is nothing but
the solution of the given equation.
As a consequence of this Principle terms can be transposed from one side (member) of the
equation to the other by simply changing their sign.
Example: In the equation x-7=3, if we transpose 7 to the right hand side of the equation we
get x=3+7 or x=10.
The equation 9y-7=8y-4 can be easily solved if we first transpose terms as follows:
9y-8y=-4+7, from which we get y=3. To check if y=3 is in fact the solution of the given
equation, we substitute this value of y in the equation. Doing this we get: 9(3)-7=8(3)-4 or
27-7=24-4 or 20=20. Since this is a true statement, y=3 is the solution of the equation.
Multiplication Principle of Equality: If we multiply (or divide) both sides of an equation
by the same non-zero number or expression we get an equivalent equation.
Examples: Given the equation
5  x 2  x 17 x  4


3
15
30
if we multiply both sides by 30 we get
10(5  x)  2(2  x)  17 x  4
We can verify that x=2 is the solution of both equations, therefore they are equivalent.
Using the Distributive Law of multiplication to remove parentheses in the second equation
we get,
50-10x-4+2x=17x-4
-10x+2x-17x=-4+4-50
-25x=-50
Finally if we divide both sides of this equation by -25 we get the solution, x=2.
As an immediate consequence of this principle we have:
i.
If a term is multiplying (dividing) one side of the equation it can be transposed to the
other side dividing (multiplying).
ii.
To remove denominators in an equation we multiply both sides of the equation by the
least common denominator (lcd) of all denominators.
SOLVING LINEAR EQUATIONS .SUMMARY: To solve a linear equation (first degree
polynomial equation) in one variable (unknown) we proceed as follows:
1.
2.
3.
4.
Remove denominators (if they exist)
Remove parentheses (if they exist)
Simplify as much as possible both sides of the equation.
Transpose like terms in such a way that the variable terms end up on one side and the
independent terms (constants) on the other side.
5. Again, simplify as much as possible both sides of the equation.
6. Use the Multiplication Principle of Equality to isolate the unknown.
7. Verify the result.
Example: Solve the equation
9 x 1 x x
  
7
2 9 18
Multiplying both sides of the equation by the lcd, 7x18=126 we get:
9  x 1 
x x 
126 
   126   
2
 7
 9 18 
18  9  x   63  14 x  7 x
Removing parentheses (distributive law) we get:
162 18x  63  14x  7 x
Combining like terms on both sides we get:
18x  225  7 x
Transposing like terms (this time, for simplicity, we will transpose the variable terms to the
right hand side) we get:
225  7 x 18x
Combining like terms:
225  25x
Finally, dividing both sides by 25:
9 x
To verify if x=9 is the solution we substitute this value of x into the given equation:
99 1 9 9
  
7
2 9 18
0
1
9
 1
2
18
1 9
1 1

or

2 18
2 2
which is a true statement.
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EXERCISES
Solve for x equations 1-4:
1. 3 x  2( x  2)  10  2 x
x x
  8
3 12
x  2 5x  4
7x
3.

 1
15
30
45
2
4. 3( x  1)  5( x  1)  3( x  1) 2  12
2.
5. Solve for l the equation P  2l  2w. What is the value of l when P  28 and w  4?
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