2.4 Use a general strategy to solve linear equations  (Page 2/6)

Solve: $10\left[3-8\left(2s-5\right)\right]=15\left(40-5s\right)$ .

Solution

Simplify from the innermost parentheses first.
Combine like terms in the brackets.
Distribute.
Add $160s$ to get the s’s to the right.
Simplify.
Subtract 600 to get the constants to the left.
Simplify.
Divide.
Simplify.
Check:
Substitute $s=\mathrm{-2}$ .

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Solve: $0.36(100n+5)=0.6(30n+15)$ .

Solution

Distribute.
Subtract $18n$ to get the variables to the left.
Simplify.
Subtract $1.8$ to get the constants to the right.
Simplify.
Divide.
Simplify.
Check:
Let $n=0.4$ .

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Classify equations

Consider the equation we solved at the start of the last section, $7x+8=\mathrm{-13}$ . The solution we found was $x=\mathrm{-3}$ . This means the equation $7x+8=\mathrm{-13}$ is true when we replace the variable, x , with the value $\mathrm{-3}$ . We showed this when we checked the solution $x=\mathrm{-3}$ and evaluated $7x+8=\mathrm{-13}$ for $x=\mathrm{-3}$ .

If we evaluate $7x+8$ for a different value of x , the left side will not be $\mathrm{-13}$ .

The equation $7x+8=\mathrm{-13}$ is true when we replace the variable, x , with the value $\mathrm{-3}$ , but not true when we replace x with any other value. Whether or not the equation $7x+8=\mathrm{-13}$ is true depends on the value of the variable. Equations like this are called conditional equations.

All the equations we have solved so far are conditional equations.

Now let’s consider the equation $2y+6=2\left(y+3\right)$ . Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y .

Distribute.
Subtract $2y$ to get the $y$ ’s to one side.
Simplify—the $y$ ’s are gone!

But $6=6$ is true.

This means that the equation $2y+6=2\left(y+3\right)$ is true for any value of y . We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable like this is called an identity    .

What happens when we solve the equation $5z=5z-1$ ?

Subtract $5z$ to get the constant alone on the right.
Simplify—the $z$ ’s are gone!

But $0\ne \text{−}1$ .

Solving the equation $5z=5z-1$ led to the false statement $0=\mathrm{-1}$ . The equation $5z=5z-1$ will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction    .

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

$6(2n-1)+3=2n-8+5(2n+1)$

Solution

Distribute.
Combine like terms.
Subtract $12n$ to get the $n$ ’s to one side.
Simplify.
This is a true statement.	The equation is an identity. The solution is all real numbers.

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