The Addition Principle of Equality

Download Report

Transcript The Addition Principle of Equality

2.2
The Addition Principle of Equality
1. Determine whether a given equation is linear.
2. Solve linear equations in one variable using the
addition principle.
3. Solve equations with variables on both sides of
the equal sign.
4. Solve identities and contradictions.
5. Solve application problems.
Addition Principle of Equality
Linear equation: An equation in which each
variable term contains a single variable raised
to an exponent of 1.
3x  5  12
a5
3x  2y  7
Nonlinear equations: 2x 2  3x  5  7
3x 3  5  12
Linear equation in one variable: An equation
that can be written in the form ax + b = c,
where a, b, and c are real numbers and a  0.
Addition Principle of Equality
5
Left side
= 5
5 = 5+2
5+2 = 5+2
Right side
We can add (or subtract) the same quantity to (from)
both sides of an equation and still have a true
statement.
Addition Principle of Equality
If a = b, then a + c = b + c
for all real numbers a, b, and c.
To clear a term in an equation, add the additive
inverse of that term to both sides of the equation.
5
Additive Inverse
–5
–3
3
2x
–7x
–2x
7x
x + 15 is an expression,
so we can’t use the
addition principle of
equality!!!!
x = 3 is the same as 3 = x.
Addition Principle of Equality
Goal: x = some number
x – 19 = –34
+ 19 + 19
x + 0 = –15
x = 15
Check:
Additive Inverse: 19
Since we added 19 to
the left side, we must
add 19 to the right side
as well.
x 19  34
15 19  34
34  34
Replace x with –15 .
True, so –15 is the solution.
Addition Principle of Equality
Goal: x = some number
x + 7 = 20
Additive Inverse: –7
–7 –7
x + 0 = 13
x = 13
Check:
x + 7 = 20
13 + 7 = 20
20 = 20
True, so 13 is the solution.
Addition Principle of Equality
Goal: m = some number
m + 5 = 12 – 4
Simplify sides if you can.
m+5=8
–5 –5
m+0=3
Additive Inverse: –5
m=3
Check: m + 5 = 12 – 4
3 + 5 = 12 – 4
8=8
True, so 3 is the solution.
Addition Principle of Equality
To Solve Linear Equations
1. Simplify both sides of the equation as needed.
a. Distribute to clear parentheses.
b. Combine like terms.
2. Use the addition principle so that all variable terms
are on one side of the equation.
3. Use the addition principle so that all constants are
on the other side.
Addition Principle of Equality
Solve: 3x  5  2x  4
Additive Inverse: –2x
–2x
–2x
1x  5  0x  4
x54
–5 –5
x  0  1
x  1
Additive Inverse: –5
Check: 3 1  5  2 1  4
 3  5  2  4
x  1 is the solution.
22
Addition Principle of Equality
Solve: 4 x  2x  3  x  7
2x  3  x  7
–x
–x
1x  3  0  7
x  3  7
–3 –3
Simplify sides if you can.
Additive Inverse: –x
Additive Inverse: –3
x  10
Check:
4 10  2 10  3   10  7
x  10 is the solution.
 40  20  3  17
 17  17
Addition Principle of Equality
Solve: 2 x  4   x  10
2x  8  x  10
–x
–x
x  8  10
+8 +8
x  18
Simplify sides if you can.
Additive Inverse: –x
Additive Inverse: +8
Check: 218  4   18  10
214   18  10
28  28
x  18 is the solution.
Solve for t.
a) t = 5
b) t = 3
c) t = 1
d) t = –1
2.2
– 8t + 4 +14t = – 4t + 3 + 9t
Solve for t.
a) t = 5
b) t = 3
c) t = 1
d) t = –1
2.2
– 8t + 4 +14t = – 4t + 3 + 9t
Addition Principle of Equality
Solve: 3 x  4   2x  x  12
3x  12  2x  x  12
3x  12  3x  12
Simplify sides if you can.
Additive Inverse: –3x
–3x
–3x
0x  12  0x  12
 12  12
Lost the variable term!!
True statement.
Identity
Solution is ALL REAL NUMBERS.
Identity: An equation that has every real number as a
solution.
Addition Principle of Equality
Solve: 2 x  4   3 x  1  x Simplify sides if you can.
2x  8  3x  3  x
Additive Inverse: –2x
2x  8  2x  3
–2x
–2x
0x  8  0x  3
Lost the variable term!!
False statement.
8  3
Contradiction
There is NO SOLUTION.
 
 
Contradiction: An equation that has no real number
solution
Addition Principle of Equality
Solve:
55x  3  64 x  2  12  15
25x  15  24x  12  3
x  3  3
+3 +3
x0
Additive Inverse: +3
Check:
550  3  640  2  12  15
5 3  6 2  12  15
x  0 is the solution.
 15  12  12  15
 3  3