How To Solve 3 7

Commencement-Caste EQUATIONS AND INEQUALITIES

In this chapter, we will develop sure techniques that assistance solve problems stated in words. These techniques involve rewriting problems in the form of symbols. For case, the stated trouble

"Notice a number which, when added to three, yields 7"

may be written every bit:

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

so on, where the symbols ?, n, and x correspond the number nosotros desire to find. We telephone call such shorthand versions of stated problems equations, or symbolic sentences. Equations such every bit 10 + three = seven are first-degree equations, since the variable has an exponent of i. The terms to the left of an equals sign make up the left-manus fellow member of the equation; those to the right brand upwardly the right-hand fellow member. Thus, in the equation 10 + 3 = 7, the left-hand member is x + 3 and the right-mitt member is 7.

SOLVING EQUATIONS

Equations may be truthful or false, merely every bit word sentences may be truthful or false. The equation:

3 + x = seven

will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (iv in this example) is called the solution of the equation. We can determine whether or not a given number is a solution of a given equation by substituting the number in place of the variable and determining the truth or falsity of the result.

Instance 1 Determine if the value 3 is a solution of the equation

4x - 2 = 3x + 1

Solution We substitute the value 3 for x in the equation and see if the left-manus fellow member equals the right-hand member.

iv(3) - 2 = 3(3) + ane

12 - 2 = 9 + 1

10 = ten

Ans. 3 is a solution.

The commencement-degree equations that we consider in this affiliate accept at near one solution. The solutions to many such equations can be determined by inspection.

Case ii Find the solution of each equation past inspection.

a. x + 5 = 12
b. 4 · x = -20

Solutions a. 7 is the solution since 7 + five = 12.
b. -five is the solution since 4(-5) = -20.

SOLVING EQUATIONS USING ADDITION AND SUBTRACTION PROPERTIES

In Section 3.1 nosotros solved some uncomplicated beginning-degree equations by inspection. However, the solutions of most equations are not immediately evident past inspection. Hence, we demand some mathematical "tools" for solving equations.

EQUIVALENT EQUATIONS

Equivalent equations are equations that have identical solutions. Thus,

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

are equivalent equations, considering v is the only solution of each of them. Notice in the equation 3x + 3 = x + 13, the solution five is not evident past inspection merely in the equation 10 = 5, the solution v is evident by inspection. In solving whatsoever equation, we transform a given equation whose solution may non be obvious to an equivalent equation whose solution is easily noted.

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

If the same 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 1 Write an equation equivalent to

x + 3 = vii

past subtracting 3 from each fellow member.

Solution Subtracting iii from each fellow member yields

x + 3 - 3 = 7 - 3

or

ten = 4

Observe that x + 3 = 7 and ten = 4 are equivalent equations since the solution is the same for both, namely 4. The next example shows how we tin can generate equivalent equations by first simplifying one or both members of an equation.

Example two Write an equation equivalent to

4x- 2-3x = iv + 6

past combining like terms and so past adding 2 to each member.

Combining like terms yields

x - ii = ten

Adding ii to each member yields

10-2+2 =ten+2

x = 12

To solve an equation, we use the improver-subtraction property to transform a given equation to an equivalent equation of the form x = a, from which we tin find the solution by inspection.

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

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

2x + 1- 1 = x - ii- 1

2x = ten - 3

If we now add -ten to (or subtract x from) each fellow member, we get

2x-x = x - 3 - x

x = -3

where the solution -3 is obvious.

The solution of the original equation is the number -iii; however, the answer is often displayed in the grade of the equation x = -iii.

Since each equation obtained in the process is equivalent to the original equation, -iii is also a solution of 2x + 1 = 10 - 2. In the above instance, we can check the solution by substituting - 3 for ten in the original equation

ii(-3) + i = (-three) - 2

-five = -v

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

If a = b then b = a

This enables u.s.a. to interchange the members of an equation whenever nosotros please without having to exist concerned with whatsoever changes of sign. Thus,

If 4 = x + 2 then 10 + 2 = 4

If x + 3 = 2x - 5 then 2x - 5 = x + 3

If d = rt so rt = d

In that location may be several unlike ways to apply the addition property above. Sometimes ane method is amend than some other, and in some cases, the symmetric property of equality is as well helpful.

Example 4 Solve 2x = 3x - nine. (1)

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

2x - 3x = 3x - nine - 3x

-x = -ix

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

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

ix = x

from which the solution 9 is obvious. If we wish, we can write the concluding equation as ten = 9 by the symmetric holding of equality.

SOLVING EQUATIONS USING THE Sectionalisation PROPERTY

Consider the equation

3x = 12

The solution to this equation is 4. Also, note that if we divide each member of the equation by 3, nosotros obtain the equations

whose solution is besides four. In general, nosotros accept the following holding, which is sometimes called the partition 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 ane Write an equation equivalent to

-4x = 12

by dividing each member past -iv.

Solution Dividing both members by -4 yields

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

Instance 2 Solve 3y + 2y = xx.

Nosotros first combine similar terms to go

5y = 20

And then, dividing each member by 5, we obtain

In the next case, nosotros use the addition-subtraction property and the division property to solve an equation.

Example iii Solve 4x + vii = x - 2.

Solution Start, nosotros add together -x and -7 to each member to become

4x + 7 - x - 7 = x - two - x - i

Next, combining like terms yields

3x = -9

Concluding, we divide each member by three to obtain

SOLVING EQUATIONS USING THE MULTIPLICATION PROPERTY

Consider the equation

The solution to this equation is 12. Also, note that if nosotros multiply each member of the equation by iv, we obtain the equations

whose solution is also 12. In general, nosotros have the following property, 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 1 Write an equivalent equation to

by multiplying each member by half-dozen.

Solution Multiplying each member by 6 yields

In solving equations, nosotros apply the above belongings to produce equivalent equations that are costless of fractions.

Case 2 Solve

Solution First, multiply each fellow member by 5 to get

Now, divide each member by 3,

Example 3 Solve .

Solution Beginning, simplify above the fraction bar to become

Next, multiply each member by iii to obtain

Last, dividing each fellow member by 5 yields

Farther SOLUTIONS OF EQUATIONS

Now we know all the techniques needed to solve almost first-degree equations. In that location is no specific lodge in which the properties should be applied. Any 1 or more of the following steps listed on folio 102 may be appropriate.

Steps to solve first-degree equations:

1. Combine like terms in each member of an equation.
2. Using the add-on 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 member.
4. Utilize the multiplication property to remove fractions.
5. Use the sectionalization property to obtain a coefficient of ane for the variable.

Case ane Solve 5x - 7 = 2x - 4x + xiv.

Solution Offset, we combine like terms, 2x - 4x, to yield

5x - 7 = -2x + xiv

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

5x - 7 + 2x + seven = -2x + xiv + 2x + 1

7x = 21

Finally, we split up each member by 7 to obtain

In the next instance, we simplify higher up the fraction bar earlier applying the properties that we take been studying.

Example 2 Solve

Solution Start, we combine like terms, 4x - 2x, to go

So we add together -three to each member and simplify

Next, we multiply each member by iii to obtain

Finally, nosotros divide each member by ii to go

SOLVING FORMULAS

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

Instance 1 In the formula d = rt, find 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) = (3)t

eight = t

It is often necessary to solve formulas or equations in which there is more than than one variable for one of the variables in terms of the others. We employ 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 past dividing both members by r to yield

from which, by the symmetric law,

In the above instance, we solved for t past applying the division property to generate an equivalent equation. Sometimes, it is necessary to apply more than one such holding.

Example 3 In the equation ax + b = c, solve for 10 in terms of a, b and c.

Solution We tin solve for x by first adding -b to each member to get

and then dividing each fellow member past a, we have

How To Solve 3 7,

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