4 7 3x 1 16

FIRST-Degree EQUATIONS AND INEQUALITIES

In this chapter, we volition develop certain techniques that aid solve problems stated in words. These techniques involve rewriting problems in the form of symbols. For example, the stated problem

"Observe a number which, when added to 3, yields 7"

may be written equally:

3 + ? = 7, 3 + n = seven, 3 + ten = 1

and so on, where the symbols ?, n, and x correspond the number we want to find. We call such shorthand versions of stated problems equations, or symbolic sentences. Equations such as x + 3 = vii are first-caste equations, since the variable has an exponent of i. The terms to the left of an equals sign make up the left-hand fellow member of the equation; those to the right brand up the right-hand fellow member. Thus, in the equation x + 3 = 7, the left-hand member is x + 3 and the right-manus member is 7.

SOLVING EQUATIONS

Equations may exist true or false, only as word sentences may be truthful or imitation. The equation:

iii + x = 7

will be simulated if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (4 in this instance) is chosen the solution of the equation. Nosotros can determine whether or non 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.

Case ane Make up one's mind if the value three is a solution of the equation

4x - two = 3x + one

Solution We substitute the value 3 for 10 in the equation and see if the left-hand member equals the right-hand fellow member.

iv(iii) - ii = 3(3) + 1

12 - ii = 9 + 1

10 = 10

Ans. iii is a solution.

The outset-degree equations that we consider in this chapter have at most one solution. The solutions to many such equations can exist adamant by inspection.

Case 2 Find the solution of each equation by inspection.

a. x + v = 12
b. 4 · 10 = -twenty

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

SOLVING EQUATIONS USING Improver AND SUBTRACTION Backdrop

In Department 3.1 nosotros solved some simple first-degree equations by inspection. Yet, the solutions of most equations are not immediately axiomatic by inspection. Hence, we need some mathematical "tools" for solving equations.

EQUIVALENT EQUATIONS

Equivalent equations are equations that have identical solutions. Thus,

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

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

The following property, sometimes called the addition-subtraction belongings, is i way that nosotros tin 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.

Instance 1 Write an equation equivalent to

x + 3 = 7

past subtracting three from each fellow member.

Solution Subtracting 3 from each member yields

ten + 3 - 3 = 7 - iii

10 = iv

Detect that x + three = vii and 10 = four are equivalent equations since the solution is the same for both, namely iv. The next example shows how we tin can generate equivalent equations by first simplifying one or both members of an equation.

Example 2 Write an equation equivalent to

4x- 2-3x = four + half-dozen

by combining similar terms and so by calculation 2 to each member.

Combining like terms yields

10 - 2 = x

Adding two to each member yields

x-2+2 =x+ii

10 = 12

To solve an equation, we apply the add-on-subtraction property to transform a given equation to an equivalent equation of the form x = a, from which we can find the solution past inspection.

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

Nosotros want to obtain an equivalent equation in which all terms containing x are in ane member and all terms not containing x are in the other. If we first add -one to (or subtract one from) each fellow member, we get

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

2x = x - three

If nosotros now add -x to (or decrease x from) each member, nosotros go

2x-10 = x - 3 - x

10 = -three

where the solution -three is obvious.

The solution of the original equation is the number -3; all the same, the reply is often displayed in the class of the equation ten = -3.

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

2(-3) + one = (-3) - 2

-5 = -5

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

If a = b and then b = a

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

If 4 = x + ii then 10 + ii = four

If x + iii = 2x - v and then 2x - v = 10 + 3

If d = rt then rt = d

There may be several different ways to apply the addition holding above. Sometimes one method is meliorate than another, and in some cases, the symmetric property of equality is also helpful.

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

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

2x - 3x = 3x - 9 - 3x

-x = -ix

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

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

9 = ten

from which the solution 9 is obvious. If we wish, nosotros can write the concluding equation as 10 = 9 by the symmetric property of equality.

SOLVING EQUATIONS USING THE Sectionalisation Belongings

Consider the equation

3x = 12

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

whose solution is also 4. In general, nosotros have the following property, which is sometimes chosen the division belongings.

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

In symbols,

are equivalent equations.

Example i Write an equation equivalent to

-4x = 12

by dividing each member past -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.

Example 2 Solve 3y + 2y = 20.

We get-go combine similar terms to get

5y = xx

Then, dividing each member by 5, we obtain

In the side by side example, we utilise the addition-subtraction property and the division property to solve an equation.

Example 3 Solve 4x + 7 = 10 - 2.

Solution Beginning, we add -x and -7 to each member to get

4x + 7 - ten - 7 = x - 2 - x - ane

Next, combining similar terms yields

3x = -ix

Last, we carve up each fellow member by 3 to obtain

SOLVING EQUATIONS USING THE MULTIPLICATION Holding

Consider the equation

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

whose solution is also 12. In general, we have the following belongings, which is sometimes called the multiplication property.

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.

Instance i Write an equivalent equation to

past multiplying each member by 6.

Solution Multiplying each member by 6 yields

In solving equations, we utilize the higher up property to produce equivalent equations that are costless of fractions.

Example ii Solve

Solution Starting time, multiply each member by 5 to get

Now, divide each member past 3,

Case iii Solve .

Solution First, simplify above the fraction bar to get

Next, multiply each member past 3 to obtain

Last, dividing each member by 5 yields

FURTHER SOLUTIONS OF EQUATIONS

Now we know all the techniques needed to solve about offset-degree equations. There is no specific guild in which the backdrop should be practical. Whatsoever one or more of the post-obit steps listed on page 102 may exist appropriate.

Steps to solve showtime-caste equations:

Combine like terms in each member of an equation.
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.
Combine similar terms in each member.
Use the multiplication property to remove fractions.
Use the partition property to obtain a coefficient of 1 for the variable.

Example 1 Solve 5x - 7 = 2x - 4x + fourteen.

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

5x - 7 = -2x + fourteen

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

5x - seven + 2x + 7 = -2x + 14 + 2x + one

7x = 21

Finally, we divide each member by vii to obtain

In the side by side case, we simplify above the fraction bar before applying the properties that we accept been studying.

Example 2 Solve

Solution Commencement, we combine like terms, 4x - 2x, to get

And then we add -3 to each member and simplify

Adjacent, nosotros multiply each fellow member past 3 to obtain

Finally, we divide each fellow member by 2 to get

SOLVING FORMULAS

Equations that involve variables for the measures of two or more 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 nosotros used in the preceding sections.

Example ane In the formula d = rt, find t if d = 24 and r = iii.

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

d = rt

(24) = (iii)t

eight = t

It is often necessary to solve formulas or equations in which there is more than one variable for one of the variables in terms of the others. We utilize the same methods demonstrated in the preceding sections.

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

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

from which, past the symmetric law,

In the above example, we solved for t by applying the sectionalization property to generate an equivalent equation. Sometimes, it is necessary to utilize more than one such property.

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

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