# 5.2: Solving Equations

- Page ID
- 49367

## Types of Equations

**Identity**

Some equations are always true. These equations are called identities. **Identities** are equations that are true for all acceptable values of the variable, that is, for all values in the domain of the equation.

\(5x=5x\) is true for all acceptable values of \(x\).

\(y+1=y+1\) is true for all acceptable values of \(y\).

\(2+5=7\) is true, and no substitutions are necessary.

**Contradiction**

Some equations are never true. These equations are called contradictions. **Contradictions** are equations that are never true regardless of the value substituted for the variable.

\(x=x+1\) is never true for any acceptable value of \(x\).

\(0 ⋅ k=14\) is never true for any acceptable value of \(k\).

\(2=1\) is never true.

**Conditional ****Equation**

The truth of some equations is conditional upon the value chosen for the variable. Such equations are called conditional equations. **Conditional equations** are equations that are true for at least one replacement of the variable and false for at least one replacement of the variable.

\(x+6=11\) is true only on the condition that \(x=5\).

\(y−7=−1\) is true only on the condition that \y=6\).

## Solutions and Equivalent Equations

The collection of values that make an equation true are called **solutions** of the equation. An equation is **solved** when all its solutions have been found.

**Equivalent ****Equations**

Some equations have precisely the same collection of solutions. Such equations are called **equivalent equations**. The equations

\(2x+1=7, 2x=6\) and \(x=3\)

are equivalent equations because the only value that makes each one true is 3.

## Sample Set A

Tell why each equation is an identity, a contradiction, or conditional.

The equation \(x-4 = 6\) is a conditional equation since it will be true only on the condition that \(x = 10\)

The equation \(x−2=x−2\) is an identity since it is true for all values of \(x\). For example,

if \(x = 5, 5-2 = 5-2\) is true

\(x = -7, -7-2 = -7-2\) is true

The equation \(a+5=a+1\) is a contradiction since every value of a produces a false statement. For example,

if \(a = 8, 8 + 5 = 8 + 1\) is false

if \(a = -2, -2 + 5 = -2 + 1\) is false

## Practice Set A

For each of the following equations, write "identity," "contradiction," or "conditional." If you can, find the solution by making an educated guess based on your knowledge of arithmetic.

\(x+1=10\)

**Answer**-
conditional, \(x = 9\)

\(y−4=7\)

**Answer**-
conditional, \(y = 11\)

\(5a=25\)

**Answer**-
conditional, \(a = 5\)

\(\dfrac{x}{4} = 9\)

**Answer**-
conditional, \(x = 36\)

\(\dfrac{18}{b} = 6\)

**Answer**-
conditional, \(b = 3\)

\(y−2=y−2\)

**Answer**-
identity

\(x+4=x−3\)

**Answer**-
contradiction

\(x+x+x=3x\)

**Answer**-
identity

\(8x=0\)

**Answer**-
conditional, \(x = 0\)

\(m−7=−5\)

**Answer**-
conditional, \(m = 2\)

## Literal Equations

The ideal gas law is easy to remember and apply in solving problems, as long as you get the **proper values a**

An equation is solved for a particular variable if that variable alone equals an expression that does not contain that particular variable.

The following equations are examples of literal equations.

- \(y=2x+7\). It is solved for \(y\).
- \(d=rt\). It is solved for \(d\).
- \(I=prt\). It is solved for \(I\).
- \(z = \dfrac{x-u}{s}\). It is solved for \(z\).
- \(y+1=x+4\). This equation is not solved for any particular variable since no variable is isolated.

## Solving Equation of the form \(x+a=b\) and \(x−a=b\)

Recall that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side.

\(\begin{array}{ccc}

\text { This is the } & \text { this } \\

\text { number } & \text { same as } & \text { number } \\

\downarrow & \downarrow & \downarrow \\

x & = & 6 \\

x+2 & = & 8 \\

x-1 & = & 5

\end{array}\)

**This suggests the following procedures:**

- We can obtain an equivalent equation (an equation having the same solutions as the original equation) by
**adding**the**same number**to**both sides**of the equation. - We can obtain an equivalent equation by
**subtracting**the**same number**from**both sides**of the equation.

We can use these results to isolate x, thus solving for x.

\(\begin{array}{flushleft}

x+a&=&b & \text{ The } a \text{ is associated with } x \text{ by addition. Undo the association} \\

x + a - a&=&b - a & \text{ by subtracting } a \text{ from both sides} \\

x + 0&=&b - a & a - a = 0 \text{ and } 0 \text{ is the additive identity. } x + 0 = x. \\

x&=&b - a & \text{ this equation is equivalent to the first equation, and it is solved for } x

\end{array}\)

\(\begin{array}{flushleft}

x-a&=&b & \text{ The } a \text{ is associated with } x \text{ by subtraction. Undo the association} \\

x - a + a&=&b + a & \text{ by adding } a \text{ to both sides} \\

x + 0&=&b + a &-a + a = 0 \text{ and } 0 \text{ is the additive identity. } x + 0 = x. \\

x&=&b + a & \text{ this equation is equivalent to the first equation, and it is solved for } x

\end{array}\)

To solve the equation \(x+a=b\) for \(x\), subtract a from both sides of the equation.

To solve the equation \(x−a=b\) for \(x\), add a to both sides of the equation.

## Sample Set B

Solve \(x + 7 = 10\) for \(x\)

\(\begin{array}{flushleft}

x+7&=&10 &7 \text{ is associated with } x \text{ by addition. Undo the association }\\

x+7 - 7&=&10 - 7 &\text{ by subtracting } 7 \text{ from both sides }\\

x+0&=&3 &7-7=0 \text{ and } 0 \text{ is the additive identity. } x + 0 = x\\

x&=&3 &x \text{ is isolated, and the equation } x + 7 = 10. \\

&&&\text{Therefore, these two equation have the same solution.}\\

&&&\text{The solution to } x = 3 \text{ is clearly } 3.\\

&&&\text{Thus, the solution to } x + 7 = 10 \text{ is also } 3.

\end{array}\)

**Check:** Substitute \(3\) for \(x\) in the original equation.

\(\begin{array}{flushleft}

x + 7&=&10\\

3+7&=&10&\text{Is this correct?}\\

10&=&10&\text{Yes, this is correct}

\end{array}\)

Solve \(m - 2 = -9\) for \(m\)

\(\begin{array}{flushleft}

m-2&=-9&2\text{ is associated with } m \text{ by subtraction. Undo the association}\\

m-2+2&=-9+2&\text{ by adding } 2 \text{ from both sides. }\\

m+0&=-7&-2+2=0 \text{ and } 0 \text{ is the additive identity. }m+0=m.\\

m&=-7

\end{array}\)

**Check: **Substitute \(-7\) for \(m\) in the original equation.

\(\begin{array}{flushleft}

m-2&=&-9\\

-7-2&=&-9&\text{ Is this correct? }\\

-9&=&-9&\text{ Yes, this is correct.}

\end{array}\)

Solve \(y - 2.181 = -16.915\) for \(y\).

\(\begin{array}{flushleft}

y-2.181&=&-16.915\\

y-2.181+2.181&=&-16.915+2.181\\

y&=&-14.734\\

\end{array}\)

Solve \(y + m = s\) for \(y\)

\(\begin{array}{flushleft}

y+m&=&s&m\text{ is associated with } y { by addition. Undo the association }\\

y+m-m&=&s-m&\text{ by subtracting } m \text{ from both sides }\\

y+0&=&s-m&m-m=0\text{ and } 0 \text{ is the additive identity. }y+0=y.\\

y&=&s-m

\end{array}\)

**Check:** Substitute \(s-m\) for \(y\) in the original equation.

\(\begin{array}{flushleft}

y+m&=&s\\

s-m+m&=&s&\text{ Is this correct? }\\

s&=&s&\text{ True. Yes, this is correct.}

\end{array}\)

Solve \(k-3h=-8h+5\) for \(k\).

\(\begin{array}{flushleft}

k-3h&=&8h+5&3h\text{ is associated with } k \text{ by subtraction. Undo the association }\\

k-3h+3h&=&-8h+5+3h&\text{ by adding } 3h \text{ to both sides. }\\

k+0&=&-5h+5&-3h+3h=0 \text{ and } 0 \text{ is the additive identity. }k+0=k.\\

k&=&-5h+5

\end{array}\)

## Practice Set B

Solve \(y−3=8\) for \(y\).

**Answer**-
\(y = 11\)

Solve \(x + 9 = -4\) for \(x\)

**Answer**-
\(x = -13\)

Solve \(m + 6 = 0\) for \(m\)

**Answer**-
\(m = -6\)

Solve (g - 7.2 = 1.3\) for \(g\)

**Answer**-
\(g = 8.5\)

Solve \(f + 2d = 5d\) for \(f\).

**Answer**-
\(f = 3d\)

Solve \(x + 8y = 2y - 1\) for \(x\)

**Answer**-
\(x = -6y - 1\)

Solve \(y + 4x - 1 = 5x + 8\) for \(y\).

**Answer**-
\(y = x + 9\)

## Exercises

For the following problems, classify each of the equations as an identity, contradiction, or conditional equation.

\(m+6=15\)

**Answer**-
conditional

\(y−8=−12\)

\(x+1=x+1\)

**Answer**-
identity

\(k−2=k−3\)

\(g+g+g+g=4g\)

**Answer**-
identity

\(x+1=0\)

For the following problems, determine which of the literal equations have been solved for a variable. Write "solved" or "not solved."

\(y=3x+7\)

**Answer**-
solved

\(m=2k+n−1\)

\(4a=y−6\)

**Answer**-
not solved

\(hk=2k+h\)

\(2a=a+1\)

**Answer**-
not solved

\(5m=2m−7\)

\(m=m\)

**Answer**-
not solved

For the following problems, solve each of the conditional equations.

\(h−8=14\)

\(k+10=1\)

**Answer**-
\(k = -9\)

\(m−2=5\)

\(y+6=−11\)

**Answer**-
\(y=−17\)

\(y−8=−1\)

\(x+14=0\)

**Answer**-
\(x=−14\)

\(m−12=0\)

\(g+164=−123\)

**Answer**-
\(g=−287\)

\(h−265=−547\)

\(x+17=−426\)

**Answer**-
\(x=−443\)

\(h−4.82=−3.56\)

\(y+17.003=−1.056\)

**Answer**-
\(y=−18.059\)

\(k+1.0135=−6.0032\)

Solve \(n+m=4\) for \(n\).

**Answer**-
\(n=4−m\)

Solve \(P+3Q−8=0\) for \(P\).

Solve \(a+b−3c=d−2f\) for \(b\).

**Answer**-
\(b=−a+3c+d−2f\)

Solve \(x−3y+5z+1=2y−7z+8\) for \(x\).

Solve \(4a−2b+c+11=6a−5b\) for \(c\).

**Answer**-
\(c=2a−3b−11\)

## Exercises for Review

Simplify \((4x^5y^2)^3\)

Write \(\dfrac{20x^3y^7}{5x^5y^3}\) so that only positive exponents appear.

**Answer**-
\(\dfrac{4y^4}{x^2}\)

Write the number of terms that appear in the expression \(5x^2+2x−6+(a+b)\), and then list them.

Find the product \((3x-1)^2\).

**Answer**-
\(9x^2-6x+1\)

Specify the domain of the equation \(y = \dfrac{5}{x-2}\)