Find the value of x: x+2/3-x+3/4=5-x-1/2
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FIRST-DEGREE EQUATIONS AND INEQUALITIES
In this chapter, 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 problem
"Find a number which, when added to 3, yields 7"
may be written as:
3 + ? = 7, 3 + n = 7, 3 + x = 1
and so on, where the symbols ?, n, and x represent the number we want to find. We call such shorthand versions of stated problems equations, or symbolic sentences. Equations such as x + 3 = 7 are first-degree equations, since the variable has an exponent of 1. The terms to the left of an equals sign make up the left-hand member of the equation; those to the right make up the right-hand member. Thus, in the equation x + 3 = 7, the left-hand member is x + 3 and the right-hand member is 7.
SOLVING EQUATIONS
Equations may be true or false, just as word sentences may be true or false. 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 true (4 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.
Example 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-hand member equals the right-hand member.
4(3) - 2 = 3(3) + 1
12 - 2 = 9 + 1
10 = 10
Ans. 3 is a solution.
The first-degree equations that we consider in this chapter have at most one solution. The solutions to many such equations can be determined by inspection.
Example 2 Find the solution of each equation by inspection.
a. x + 5 = 12
b. 4 · x = -20
Solutions a. 7 is the solution since 7 + 5 = 12.
b. -5 is the solution since 4(-5) = -20.
SOLVING EQUATIONS USING ADDITION AND SUBTRACTION PROPERTIES
In Section 3.1 we solved some simple first-degree equations by inspection. However, the solutions of most equations are not immediately evident 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 = x + 10, 2x = 10, and x = 5
are equivalent equations, because 5 is the only solution of each of them. Notice in the equation 3x + 3 = x + 13, the solution 5 is not evident by inspection but in the equation x = 5, the solution 5 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 way 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 + 3 - 3 = 7 - 3
or
x = 4
Notice that x + 3 = 7 and x = 4 are equivalent equations since the solution is the same for both, namely 4. The next example shows how we can generate equivalent equations by first simplifying one or both members of an equation.
Example 2 Write an equation equivalent to
4x- 2-3x = 4 + 6
by combining like terms and then by adding 2 to each member.
Combining like terms yields
x - 2 = 10
Adding 2 to each member yields
x-2+2 =10+2
x = 12
To solve an equation, we use the addition-subtraction property to transform a given equation to an equivalent equation of the form x = a, from which we can find the solution by inspection.
Example 3 Solve 2x + 1 = 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 subtract 1 from) each member, we get
2x + 1- 1 = x - 2- 1
2x = x - 3
If we now add -x to (or subtract x from) each member, we get
2x-x = x - 3 - x
x = -3
where the solution -3 is obvious.