Solve the following equations with decimals. 1.5 ( z − 2 ) + 0.6 = 0.3 z + 9 1.5\left(z-2\right)+0.6=0.3z+9

Welcome back. For this problem, we want to solve the equation 1.5 times z minus two plus 0.6 is equal to 0.3 z plus nine. Because this equation has decimals, we need to start with step zero, clearing those decimals by multiplying by a power of 10. We'll have to check each of our terms to see how many decimal places it has, and then we'll see the highest number we get and multiply by that power of 10. Let's start with our first term, 1.5 times z minus two. We have one decimal in this term, the 1.5, and it has one decimal place. Next, we have 0.6, which has one decimal place. Then on the right side, we have 0.3 z. 0.3 has one decimal place. And then we have nine, which has no decimal places. The highest number here is one, so we need to multiply by 10 to the first, which is 10, And we'll do that to both sides of the equation. We get 10 times 1.5 times z minus two plus 0.6 is equal to 10 times 0.3 z plus nine. And now we should distribute that 10 on both sides of the equation. Let's start with 10 times 1.5 times z minus two. When we multiply by 10, we move the decimal point one place to the right. So 10 times 1.5, we move the decimal point one place to the right and we get 15. So this leaves us with 15 times z minus two. Next, we'll do 10 times 0.6, which we move the decimal point one place to the right, and that gives us six. And on the right side of the equation, we have 10 times 0.3 z. We move the decimal point one place to the right, and that gives us three z. And then 10 times nine, which is 90. We have no more fractions or decimals, so step zero is all set and we move on to step one. We need to distribute our 15 on the left side of the equation. We'll do 15 times z, which is 15 z, and then 15 times minus two, which is minus 30. We still have our plus six, and on the right side, we still have three z plus 90. Next, we need to combine like terms. We have a minus 30 and a plus six on the left side of the equation, So that will combine to be minus 24. So on the left side, we have 15 z minus 24 is equal to, on the right side, three z plus 90. We don't have any more distributing or like terms that we need to worry about. So step one is all set, and we move on to step two, trying to get our variable terms to one side of the equation and our constants to the other. Currently, we have variable terms and constants on both sides of the equation, so let's try to get the variable terms to the left and the constants to the right. We'll start with the constants. We need to get rid of this minus 24 on the left side, which we can do by adding 24 to both sides of the equation. It cancels the minus 24 on the left, leaving us with 15 z. And on the right side, we get three z plus 90 plus 24, which is plus one fourteen. Now we wanna get variable terms to the left, so we need to get rid of this three z on the right, which we can do by subtracting three z from both sides of the equation. On the left side, we get 15 z minus three z, which is 12 z. We cancel three z and minus three z on the right, and that leaves us with one fourteen. Now we have variable terms at the left and constants on the right, so step two is all set and we move on to step three where we wanna use multiplication or division to isolate our variable. We need to get rid of this coefficient of 12 on the left, so we'll divide both sides of the equation by 12. When we divide the left side by 12, we have that the twelves cancel and we get z on the left side of the equation. On the right side, we get 114 divided by 12. Let's see if we can simplify that fraction, one fourteen divided by 12. It might take a little bit of work, but 114 is equal to 19 times six. And we know that 12 is equal to two times six. We can cancel a six from both the numerator and the denominator, and that's going to leave us with 19 halves. So we get the solution z is equal to 19 halves. We can also write that solution as a decimal. If we do 19 divided by two, we end up getting 9.5. So we can write our solution as z is equal to 9.5 as well. And again, 19 halves and 9.5 are the same. So this is the same solution, just written in a different way. We have our variable isolated, so that is step three all set. The last thing we need to do is plug in one form of our solution back into the original equation to verify our solution. But once that's all set, great work, you have solved this equation.

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1. Whole Numbers1h 55m

Multiplying Whole Numbers
46m

Dividing Whole Numbers
21m

Long Division
36m

Exponents
10m

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Introduction to Integers
16m

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24m

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10m

Dividing Integers
16m

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Intro To Fractions and Mixed Numbers
44m

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45m

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26m

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12m

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41m

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Ratios and Rates
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33m

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16m

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The Circle
20m

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The Distributive Property
17m

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15m

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41m

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30m

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1h 14m

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37m

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21m

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59m

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43m

Linear Inequalities in One Variable
46m

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The Rectangular Coordinate System
44m

Graph Linear Equations in Two Variables
24m

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23m

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44m

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38m

Point Slope Form
22m

Linear Inequalities in Two Variables
28m

Introduction to Relations and Functions
57m

Function Notation
15m

12. Systems of Linear Equations1h 43m

Solving Systems of Linear Equations by Graphing
41m

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15m

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45m

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Review: Evaluating Exponents
29m

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10m

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18m

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18m

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The Quotient Rule
13m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
21m

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33m

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37m

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11m

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37m

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12m

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41m

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39m

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25m

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19m

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32m

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32m

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44m

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27m

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Introduction to Radical Expressions
47m

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46m

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19m

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53m

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The Square Root Property
17m

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26m

The Quadratic Formula
40m

Graphing Quadratic Equations
30m

10. Linear Equations and Inequalities

Solving Linear Equations

Prealgebra & Beginning Algebra

10. Linear Equations and Inequalities

Solving Linear Equations

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Multiple Choice

Solve the following equations with decimals.
$1.5 (z - 2) + 0.6 = 0.3 z + 9$

$z equals 9.5$

$z equals 38$

$z = \frac{57}{9}$

$z = \frac{22}{9}$

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Solve the following equations with decimals.1.5(z−2)+0.6=0.3z+91.5\(\left\)(z-2\(\right\))+0.6=0.3z+9

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Solve the following equations with decimals.
$1.5 (z - 2) + 0.6 = 0.3 z + 9$