Download Alg 1A 2.3 - Solving Multi-Step Equations, and

DJB Lesson(key):
Alg-1A_Multi-Step Eqns(2.3)_L1
Alg 1A
2.3 - Solving Multi-Step Equations, and Clearing Equations of Decimals or Fractions
Let's solve the following slightly more
challenging equations:
We will use the same basic steps:
1) Simplify each side
> Use Distrib Prop to eliminate ( )
(1)
3x + 6 - (7x - 3) = 25
> Eliminate double-negatives
> Combine Like Terms
(2)
3(3x + 2) - 2x = 41
2) Finish with 2-step.
> Undo Addition or Subtraction
(3)
5(x-2) - 2(x+1) = 15
> Undo Multiplication or Division
3) Check our answer by substituting
We will also identify 2 types of equations where we may want to simplify the entire
equation up front, by clearing the equation completely of decimals or fractions.
Solve each equation by first simplifying each side, as applicable.
Cornell Notes format: For future practice, cover right side to hide answers; re-work problem & check .
(1)
3x + 6 - (7x - 3) = 25
3x + 6 - (7x - 3) = 25
3x + 6 - 7x - (-3) = 25
3x + 6 - 7x + 3 = 25
-4x + 9
= 25
-4x + 9 - 9
= 25 - 9
-4x
= 16
(- 4x)*(-1/4)
= 16*(-1/4)
x =-4
<< Eliminate ( ) using Distrib Prop
<< Eliminate double-negative
<< Combine Lik e Terms
<< subtract 9 on both sides
<< simplify
<< divide by -4 on both sides
Check by substituting into orig. eqn.
3(-4) + 6 - [7(-4) - 3] = 25
(2)
3(3x + 2) - 2x = 41
3(3x + 2) - 2x = 41
3*3x + 3*2 - 2x = 41
9x + 6 - 2x = 41
7x + 6
= 41
7x + 6 - 6 = 41 - 6
7x
= 35
(1/7)*7x = 35*(1/7)
x
=5
<< simplify to show this is true
<< Apply Distributive Property
<< simplify
<< combine Lik e Terms
<< subtract 6 on both sides
<< simplify
<< divide by 7 on both sides
Check by substituting into orig. eqn.
3(3*5 + 2) - 2*5 = 41
(3)
5(x-2) - 2(x+1) = 15
5(x-2) - 2(x+1) = 15
<< simplify to show this is true
(3)
5(x-2) - 2(x+1) = 15
5(x-2) - 2(x+1) = 15
5x - 10 - 2x -2 = 15
3x - 12
= 15
3x - 12 + 12 = 15 + 12
3x
= 27
(1/3)*3x = 27*(1/3)
x
=9
<< Apply Distrib. Property twice
<< Combine Lik e Terms
<< add 12 to both sides
<< simplify
<< divide both sides by 3
Check by substituting into orig. eqn.
5(9 - 2) - 2(9+1) = 15
<< simplify to show this is true
Continue solving each of the following using a similar approach (1st simplify)
(4)
-3(x - 4) = 18
x=-2
(5)
3x + 2x - 5 = - 35
x=-6
(6)
(1/5)(x - 20) = 8
x = 60
(7)
2(x-5) + 6(x+3) = 40
x=4
Now let's consider 2 types of equations where we may want to simplify the entire
equation up front, by clearing the equation completely of decimals or fractions.
Taking this extra step up front will usually make the problem much easier to finish.
Clearing Decimals
If we have an equation with decimals, such as
0.4x + 0.25 = 1.85
we can make this an easier problem by first eliminating (clearing ) all of the decimals.
To clear the equation of decimals, multiply each side by the [same] power of 10
needed to move the greatest number of decimal places.
Need 1
Need 2
Need 2
0.4x + 0.25
= 1.85
100*(0.40x) + 100*(0.25) = 100*(1.85)
40x + 25
= 185
<< identify the # of decimall places
<< use the greatest power of 10
<< simplify
Alternatively, we could simply work with the decimal values without clearing them.
i.e., 1st subtracting 0.25 from both sides; then dividing both sides by 0.4
Practice
Use both methods to solve this equation:
.05x + .02(3x - 1) = 1.19
1st) working with the decimals, as is, without first clearing them; and,
2nd) by clearing the decimals up front
(should get the same answer for both)
2nd) by clearing the decimals up front
(should get the same answer for both)
Clearing Fractions
If we have an equation with fractions, such as:
(3/10)x + (1/5) = - 1/2
we can solve this by first eliminating (clearing ) all of the fractions.
To clear the fractions (denominators), multiply each side by the same number
that is a multiple of all of the denominators (e.g., LCD = LCM of the denominators)
(3/10)x + (1/5) = - 1/2
(10)*(3/10)x + 10(1/5) = 10*(-1/2)
3x
+ 2 = -5
<< identify the least common denominator
<< multiply by the LCD
<< simplify
Alternatively, we could simply work with the fractions without clearing them.
i.e., 1st subtracting 1/5 from both sides; then multiplying both sides by 10/3
Practice
Use both methods to solve this equation:
(1/6)x + (1/3)(x - 2) = 5/6
1st) working with the fractions, as is, without first clearing them; and,
2nd) by clearing the fractions up front
(should get the same answer for both)
To recap, if we opt to clear an entire equation of decimals, it can make the
values much easier to work with on paper, so it can be a useful strategy.
Similarly for clearing an entire equation of fractions.
Solve for x by clearing the decimals:
.01x + .15 = -0.5
Solve for x by first clearing the fractions (multiply by LCD):
(1/4)x + (1/8) = - (7/8)