The Solution of the Equation X 3 2 7 is

The Solution of the Equation X 3 2 7 is.

FIRST-Degree EQUATIONS AND INEQUALITIES

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

“Find a number which, when added to 3, yields 7”

may be written every bit:

3 + ? = vii, 3 + n = 7, 3 + ten = one

so on, where the symbols ?, northward, and x represent the number we want to find. Nosotros call such autograph versions of stated problems equations, or symbolic sentences. Equations such as ten + iii = 7 are first-degree equations, since the variable has an exponent of 1. The terms to the left of an equals sign make upwards the left-mitt member of the equation; those to the right make up the right-paw member. Thus, in the equation x + 3 = 7, the left-hand member is x + 3 and the correct-hand member is 7.

SOLVING EQUATIONS

Equations may exist true or false, just equally word sentences may exist true or fake. The equation:

3 + 10 = vii

will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (4 in this case) is called the solution of the equation. Nosotros 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 upshot.

Instance i
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 run across if the left-mitt member equals the right-paw member.

4(3) – 2 = 3(3) + 1

12 – 2 = nine + 1

10 = ten

Ans. 3 is a solution.

The offset-degree equations that we consider in this affiliate have at most 1 solution. The solutions to many such equations can be determined by inspection.

Instance ii
Find the solution of each equation past inspection.

a. x + five = 12

b. iv · x = -xx

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

SOLVING EQUATIONS USING ADDITION AND SUBTRACTION Backdrop

In Section 3.1 nosotros solved some simple first-caste 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 take identical solutions. Thus,

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

are equivalent equations, considering 5 is the only solution of each of them. Find in the equation 3x + 3 = x + xiii, the solution 5 is not evident by inspection simply in the equation x = five, the solution five is evident by inspection. In solving any equation, we 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 property, is one fashion that we 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 = 7

by subtracting 3 from each member.

Solution Subtracting 3 from each member yields

x + three – 3 = 7 – 3

or

x = 4

Discover that 10 + 3 = seven and x = 4 are equivalent equations since the solution is the same for both, namely 4. The side by side instance shows how we can generate equivalent equations by offset simplifying 1 or both members of an equation.

Example 2
Write an equation equivalent to

4x- ii-3x = 4 + vi

by combining like terms and and so by adding 2 to each fellow member.

Combining like terms yields

x – ii = ten

Adding two to each member yields

10-2+2 =ten+two

10 = 12

To solve an equation, we use the addition-subtraction property to transform a given equation to an equivalent equation of the form 10 = a, from which nosotros can discover the solution past inspection.

Case three
Solve 2x + one = x – 2.

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

2x + 1- 1 = x – 2- i

2x = x – 3

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

2x-x = x – iii – 10

x = -three

where the solution -3 is obvious.

The solution of the original equation is the number -3; however, the respond is often displayed in the class of the equation x = -iii.

Since each equation obtained in the procedure is equivalent to the original equation, -three is likewise a solution of 2x + 1 = x – ii. In the above example, nosotros tin can bank check the solution past substituting – 3 for x in the original equation

2(-three) + one = (-3) – two

-5 = -5

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

If a = b and then b = a

This enables u.s.a. to interchange the members of an equation whenever nosotros delight without having to be concerned with any changes of sign. Thus,

If 4 = x + 2 then ten + ii = 4

If 10 + 3 = 2x – 5 then 2x – 5 = x + 3

If d = rt then rt = d

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

Instance four
Solve 2x = 3x – 9. (1)

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

2x – 3x = 3x – 9 – 3x

-x = -9

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

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

9 = x

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

SOLVING EQUATIONS USING THE DIVISION Holding

Consider the equation

3x = 12

The solution to this equation is iv. As well, note that if we separate each fellow member of the equation by 3, nosotros obtain the equations

whose solution is too iv. In full general, we have the post-obit property, which is sometimes called the division property.

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.

Instance 1
Write an equation equivalent to

-4x = 12

by dividing each member by -4.

Solution Dividing both members past -4 yields

In solving equations, we use the in a higher place property to produce equivalent equations in which the variable has a coefficient of i.

Case 2
Solve 3y + 2y = twenty.

Nosotros kickoff combine like terms to become

5y = twenty

Then, dividing each member past 5, we obtain

In the next example, we use the add-on-subtraction holding and the segmentation holding to solve an equation.

Example 3
Solve 4x + 7 = ten – 2.

Solution First, nosotros add -x and -7 to each member to get

4x + 7 – x – 7 = x – 2 – 10 – 1

Next, combining like terms yields

3x = -9

Concluding, nosotros carve up each fellow member by 3 to obtain

SOLVING EQUATIONS USING THE MULTIPLICATION PROPERTY

Consider the equation

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

whose solution is also 12. In general, we have the post-obit property, 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.

Case 1
Write an equivalent equation to

by multiplying each member past 6.

Solution Multiplying each member by 6 yields

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

Example 2
Solve

Solution First, multiply each member by five to get

Now, divide each member by iii,

Example 3
Solve
.

Solution First, simplify above the fraction bar to get

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

Last, dividing each member by 5 yields

Farther SOLUTIONS OF EQUATIONS

Now nosotros know all the techniques needed to solve near kickoff-degree equations. In that location is no specific guild in which the properties should be applied. Any i or more of the following steps listed on page 102 may exist appropriate.

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 member.
  4. Use the multiplication property to remove fractions.
  5. Use the partition property to obtain a coefficient of 1 for the variable.

Example one
Solve 5x – vii = 2x – 4x + 14.

Solution Starting time, nosotros combine like terms, 2x – 4x, to yield

5x – 7 = -2x + 14

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

5x – 7 + 2x + vii = -2x + xiv + 2x + 1

7x = 21

Finally, we split up each member by vii to obtain

In the adjacent example, we simplify above the fraction bar earlier applying the properties that nosotros have been studying.

Example 2
Solve

Solution First, nosotros combine similar terms, 4x – 2x, to get

So we add together -iii to each fellow member and simplify

Adjacent, nosotros multiply each member by 3 to obtain

Finally, we divide each member by ii to become

SOLVING FORMULAS

Equations that involve variables for the measures of two or more than concrete quantities are called formulas. We can solve for whatever 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 nosotros used in the preceding sections.

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

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

d = rt

(24) = (3)t

viii = t

It is ofttimes necessary to solve formulas or equations in which in that location is more than 1 variable for one of the variables in terms of the others. We utilize the same methods demonstrated in the preceding sections.

Example two
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,

In the in a higher place example, we solved for t past applying the segmentation property to generate an equivalent equation. Sometimes, it is necessary to apply more than 1 such property.

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

Solution We can solve for 10 by commencement adding -b to each member to go

then dividing each member past a, we have

The Solution of the Equation X 3 2 7 is

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