In this problem, we solve the equation .
(a) Isolate the by subtracting an appropriate constant from both sides.
(b) Solve the resulting equation for .
In this problem, we solve the equation .
(a) Simplify both sides of the equation by combining like terms.
(b) Add an expression to both sides of your equation from part (a) to give an equation in which no variables are on the right-hand side.
(c) Solve the equation resulting from part (b).
(d) Check your answer! Substitute your value of into the original equation. If it doesn’t work, then do the problem again.
Solve the following equations:
(a)
(b)
(c)
(d)
Solve the following equations:
(a)
(b)
(a) Find all values of that satisfy .
(b) Find all values of that satisfy .
For what value of do the equations and have the same solution for ?
In the last section, we used addition and subtraction to solve some equations, and used multiplication and division to solve others. To solve most linear equations, however, we’ll have to use a combination of these tactics.
Solve the equation .
Solution for Problem 5.12: This equation doesn’t look exactly like any of the equations we already know how to solve. It may not be obvious immediately how to isolate . However, we can isolate by subtracting 9 from both sides:
Now we have an equation we know how to solve! We divide both sides by 8 to find .
We can check our work by substituting this value for back into our original equation. We find that , so our answer works.
We didn’t have to add first when we solved this equation. We could have divided first: We can then distribute on the left side. Since we have We then subtract from both sides of this equation to get , as before. □
The equation in Problem 5.12 is not exactly like any of the equations we solved in the previous section. However, we were still able to solve it with the same tools.
See if you can apply this strategy to the following problem.
Solve the equation .
Solution for Problem 5.13: Our first step is to simplify both sides of the equation. By grouping like terms, the left-hand side of the original equation becomes The right-hand side of the original equation becomes
Combining these results simplifies the original equation to We haven’t solved any equations in which the variable appears on both sides. We know how to handle an equation if the variable only appears on one side. So, we add to both sides to eliminate the variable from the right-hand side:
Now we have an equation we know how to solve! We add 4 to both sides to get . We then divide by 12 to find . □
We now have another strategy for solving linear equations.
Similarly, we use addition and subtraction to get all the constant terms on the other side of the equation.
Here’s a little more practice.
Solve the following equations:
(a)
(b)
(c)
(d)
Solution for Problem 5.14:
(a) Adding 13 to both sides leaves the variable term on the left while putting all the constant terms on the right: Simplifying the right-hand side gives so we now have Multiplying both sides by (which is the same as dividing both sides by 8) gives
(b) First, we use the distributive property to expand both sides: Simplifying both sides gives Next, we get all the terms with on one side of the equation and all the constants on the other side. Subtracting from both sides gives . Subtracting 9 from both sides gives . Finally, dividing both sides by 2 gives .
(c) First, make sure you see why adding 7 to doesn’t “cancel the .” This is because equals , which is . There’s still a constant term; the term is not yet isolated.
Since equals , we add to both sides of to eliminate the constant on the left side and isolate . Doing so gives us Multiplying both sides of by gives .
We could have avoided fractions entirely by multiplying both sides of by 9 on the first step to get . Since the 9’s cancel on the left side of to leave . Adding 7 to both sides gives , so , as before.
Checking our answer, we find that if , then , as required.
(d) We start by getting rid of the fractions. We eliminate the denominator on the right by multiplying both sides by 7: The 7’s on the right-hand side cancel, because So, we can write as Next, we multiply both sides by 5 to cancel the 5 in the denominator on the left-hand side: The 5’s on the left cancel, and we are left with Expanding both sides gives Simplifying both sides gives , and now we’re in familiar territory. Subtracting from both sides gives . Subtracting 28 from both sides gives . Dividing both sides by 11 gives .
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Notice that multiplying both sides of by the denominators of both fractions gave us Rather than performing these multiplications as two separate steps, we will often perform both at once. Multiplying both sides of the original equation by 5 and by 7 gives The 5 on the left cancels with the 5 in the denominator on the left, and the 7 on the right cancels with the 7 in the denominator on the right, leaving We call this process cross-multiplying.
Our last example above showed another way to simplify working with equations:
Let’s practice this strategy.
Solve the following equations:
(a)
(b)
Solution for Problem 5.15:
(a) Let’s get rid of the fractions right away. We multiply both sides of the equation by 3 to cancel the denominators that are 3, and multiply by 5 to cancel the denominators that are 5. Therefore, we can take care of both at once by multiplying by . Using the distributive property to expand, the left-hand side becomes
Multiplying the right-hand side of the original equation by 15 gives Combining this with our simplified left-hand side gives We add to both sides to get . We subtract 35 from both sides to get and divide by 28 to find .
(b) We might start by multiplying both sides by to cancel both denominators. However, since , we can cancel both denominators by multiplying both sides by 24 instead of 48: Multiplying on the left-hand side and distributing on the right gives so Dividing gives . No more fractions! Expanding the left-hand side gives us . Adding to both sides and subtracting 48 from both sides gives . Dividing by 31 gives us .
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So far, all the equations we have solved have had exactly one solution. This isn’t always the case!
(a) Find all values of that satisfy .
(b) Find all values of that satisfy .
Solution for Problem 5.16:
(a) We first simplify both sides. This gives us When we next try to get all the terms on one side by subtracting from both sides, we have
Uh-oh! What happened to the ’s? They all canceled. Worse yet, we are left with an equation that can clearly never be true, since cannot ever equal !
Since the equation can never be true, we know that the original equation can never be true either. That is, the original equation is not true for any value of . We can see why when we look back to the equation . The left-hand side is 14 greater than the right-hand side, no matter what value of we use.
We conclude that there are no solutions to the original equation.
(b) Once again, we simplify both sides of the equation, which gives Since both sides of the equation simplify to the same expression, we see that the equation is always true! No matter what value of we choose, the equation will always be true. Therefore, all values of satisfy the given equation.
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We see now that some linear equations have no solutions, and others that are satisfied by every value of the variable in the equation.
For what value of do the equations and have the same solution for ?
Solution for Problem 5.17: We know how to handle the first equation, so let’s start there. By solving the first equation for , we can find the value of that must satisfy both equations. Adding 5 to both sides of gives . Dividing by 2 then gives . This value of must also satisfy . So, when we substitute into , we must have a true equation. This substitution gives Now that we have a linear equation for , we can find . Adding 8 to both sides gives . Dividing by then gives . □
Solve the following equations:
(a)
(b)
(c)
Solve the following equations:
(a)
(b)
(c)
If , then what is the value of ?
Solve the following equations:
(a)
(b)
(c)
Solve .
Solve the following equations:
(a)
(b)
(c)