Solve 3 X 2 12

Start-DEGREE EQUATIONS AND INEQUALITIES

In this affiliate, we will develop certain techniques that help solve problems stated in words. These techniques involve rewriting problems in the form of symbols. For example, the stated trouble

"Find a number which, when added to 3, yields seven"

may be written as:

3 + ? = seven, three + n = 7, 3 + x = 1

and so on, where the symbols ?, n, and ten stand for the number we want to detect. We phone call such shorthand versions of stated issues equations, or symbolic sentences. Equations such as x + 3 = seven are first-degree equations, since the variable has an exponent of 1. The terms to the left of an equals sign make upwardly the left-hand member of the equation; those to the right brand up the right-paw member. Thus, in the equation x + 3 = 7, the left-paw member is x + 3 and the right-hand member is 7.

SOLVING EQUATIONS

Equations may be true or false, but as give-and-take sentences may be true or imitation. The equation:

3 + x = 7

will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is truthful (4 in this example) is called the solution of the equation. Nosotros tin can determine whether or not a given number is a solution of a given equation by substituting the number in identify of the variable and determining the truth or falsity of the issue.

Example 1 Determine if the value three is a solution of the equation

4x - ii = 3x + ane

Solution We substitute the value iii for x in the equation and see if the left-mitt member equals the correct-mitt member.

4(three) - 2 = 3(3) + 1

12 - ii = 9 + one

ten = x

Ans. 3 is a solution.

The first-caste equations that we consider in this chapter have at most ane solution. The solutions to many such equations can be determined by inspection.

Example ii Discover the solution of each equation by inspection.

a. ten + 5 = 12
b. 4 · x = -20

Solutions a. vii is the solution since seven + five = 12.
b. -five is the solution since 4(-five) = -twenty.

SOLVING EQUATIONS USING Improver AND SUBTRACTION Backdrop

In Section 3.1 we solved some simple first-degree equations past inspection. However, the solutions of most equations are non immediately axiomatic by inspection. Hence, we need some mathematical "tools" for solving equations.

EQUIVALENT EQUATIONS

Equivalent equations are equations that take identical solutions. Thus,

3x + 3 = 10 + 13, 3x = 10 + 10, 2x = 10, and x = 5

are equivalent equations, because five is the simply solution of each of them. Notice in the equation 3x + 3 = ten + 13, the solution 5 is not evident by inspection but in the equation 10 = 5, the solution 5 is axiomatic by inspection. In solving any equation, we transform a given equation whose solution may not exist obvious to an equivalent equation whose solution is hands noted.

The following property, sometimes called the addition-subtraction property, is one way that we can generate equivalent equations.

If the aforementioned quantity is added to or subtracted from both members of an equation, the resulting equation is equivalent to the original equation.

In symbols,

a - b, a + c = b + c, and a - c = b - c

are equivalent equations.

Example ane Write an equation equivalent to

x + 3 = 7

by subtracting 3 from each fellow member.

Solution Subtracting 3 from each member yields

x + 3 - 3 = seven - 3

or

x = 4

Notice that x + iii = 7 and x = iv are equivalent equations since the solution is the same for both, namely 4. The next case shows how we can generate equivalent equations past first simplifying one or both members of an equation.

Example 2 Write an equation equivalent to

4x- ii-3x = iv + 6

by combining like terms and then by adding 2 to each member.

Combining similar terms yields

x - ii = 10

Adding 2 to each member yields

10-2+2 =ten+2

x = 12

To solve an equation, we use the add-on-subtraction property to transform a given equation to an equivalent equation of the grade ten = a, from which we tin find the solution by inspection.

Case 3 Solve 2x + 1 = x - 2.

We want to obtain an equivalent equation in which all terms containing 10 are in ane member and all terms not containing ten are in the other. If nosotros first add together -1 to (or decrease 1 from) each member, we get

2x + i- 1 = x - 2- ane

2x = x - 3

If we at present add together -ten to (or subtract x from) each member, nosotros get

2x-x = x - 3 - x

x = -3

where the solution -3 is obvious.

The solution of the original equation is the number -3; nonetheless, the answer is frequently displayed in the form of the equation x = -three.

Since each equation obtained in the process is equivalent to the original equation, -3 is also a solution of 2x + 1 = ten - 2. In the above example, we can check the solution by substituting - 3 for x in the original equation

2(-3) + 1 = (-iii) - 2

-5 = -5

The symmetric property of equality is too helpful in the solution of equations. This property states

If a = b then b = a

This enables united states to interchange the members of an equation whenever we please without having to be concerned with any changes of sign. Thus,

If 4 = x + 2 and so x + 2 = 4

If x + 3 = 2x - five so 2x - 5 = x + iii

If d = rt then rt = d

There may be several dissimilar ways to use the improver belongings above. Sometimes i method is better than another, and in some cases, the symmetric property of equality is also helpful.

Example four Solve 2x = 3x - 9. (1)

Solution If we first add together -3x to each member, nosotros get

2x - 3x = 3x - ix - 3x

-10 = -ix

where the variable has a negative coefficient. Although nosotros can come across past inspection that the solution is 9, because -(9) = -9, we can avoid the negative coefficient by adding -2x and +9 to each member of Equation (1). In this case, nosotros get

2x-2x + ix = 3x- 9-2x+ nine

9 = x

from which the solution 9 is obvious. If we wish, we can write the last equation as x = 9 by the symmetric property of equality.

SOLVING EQUATIONS USING THE Division PROPERTY

Consider the equation

3x = 12

The solution to this equation is iv. Too, note that if we divide each member of the equation by 3, we obtain the equations

whose solution is besides 4. In general, we have the following property, which is sometimes called the division property.

If both members of an equation are divided by the aforementioned (nonzero) quantity, the resulting equation is equivalent to the original equation.

In symbols,

are equivalent equations.

Example 1 Write an equation equivalent to

-4x = 12

past dividing each member by -4.

Solution Dividing both members past -4 yields

In solving equations, we use the above property to produce equivalent equations in which the variable has a coefficient of 1.

Instance two Solve 3y + 2y = xx.

We first combine like terms to get

5y = twenty

Then, dividing each member by 5, we obtain

In the side by side case, nosotros utilize the improver-subtraction holding and the division property to solve an equation.

Case three Solve 4x + 7 = x - 2.

Solution First, we add together -10 and -7 to each member to go

4x + 7 - ten - 7 = 10 - 2 - x - one

Next, combining like terms yields

3x = -9

Concluding, we divide each member by 3 to obtain

SOLVING EQUATIONS USING THE MULTIPLICATION Belongings

Consider the equation

The solution to this equation is 12. Besides, note that if nosotros multiply each member of the equation past 4, we obtain the equations

whose solution is besides 12. In general, nosotros take the following belongings, which is sometimes called the multiplication holding.

If both members of an equation are multiplied by the same nonzero quantity, the resulting equation Is equivalent to the original equation.

In symbols,

a = b and a·c = b·c (c ≠ 0)

are equivalent equations.

Example i Write an equivalent equation to

past multiplying each member by six.

Solution Multiplying each fellow member past 6 yields

In solving equations, nosotros use the to a higher place belongings to produce equivalent equations that are costless of fractions.

Example two Solve

Solution First, multiply each member by 5 to become

Now, divide each fellow member by iii,

Instance 3 Solve .

Solution Kickoff, simplify higher up the fraction bar to get

Next, multiply each member by three to obtain

Terminal, dividing each member by five yields

FURTHER SOLUTIONS OF EQUATIONS

Now we know all the techniques needed to solve most first-degree equations. In that location is no specific order in which the properties should exist practical. Whatsoever one or more of the following steps listed on page 102 may be advisable.

Steps to solve first-degree equations:

1. Combine like terms in each member of an equation.
2. Using the addition or subtraction property, write the equation with all terms containing the unknown in one member and all terms not containing the unknown in the other.
3. Combine like terms in each fellow member.
4. Use the multiplication property to remove fractions.
5. Use the division belongings to obtain a coefficient of one for the variable.

Instance ane Solve 5x - seven = 2x - 4x + fourteen.

Solution First, we combine similar terms, 2x - 4x, to yield

5x - 7 = -2x + fourteen

Next, we add +2x and +7 to each member and combine like terms to get

5x - 7 + 2x + 7 = -2x + 14 + 2x + 1

7x = 21

Finally, we divide each member by 7 to obtain

In the next example, we simplify in a higher place the fraction bar before applying the backdrop that we have been studying.

Example 2 Solve

Solution First, we combine like terms, 4x - 2x, to become

Then we add together -3 to each fellow member and simplify

Side by side, nosotros multiply each member by 3 to obtain

Finally, we carve up each member by 2 to get

SOLVING FORMULAS

Equations that involve variables for the measures of ii or more than physical quantities are called formulas. We can solve for any one of the variables in a formula if the values of the other variables are known. We substitute the known values in the formula and solve for the unknown variable by the methods we used in the preceding sections.

Example 1 In the formula d = rt, observe t if d = 24 and r = 3.

Solution We tin can solve for t by substituting 24 for d and 3 for r. That is,

d = rt

(24) = (iii)t

8 = t

Information technology is often necessary to solve formulas or equations in which at that place is more than than one variable for 1 of the variables in terms of the others. Nosotros use the same methods demonstrated in the preceding sections.

Example 2 In the formula d = rt, solve for t in terms of r and d.

Solution We may solve for t in terms of r and d by dividing both members by r to yield

from which, by the symmetric police force,

In the in a higher place example, we solved for t by applying the sectionalisation property to generate an equivalent equation. Sometimes, it is necessary to apply more than ane such belongings.

Example three In the equation ax + b = c, solve for x in terms of a, b and c.

Solution We can solve for x by get-go calculation -b to each member to get

then dividing each member past a, nosotros have

Solve 3 X 2 12,

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