When we have an equation, we want to find the solution. The solution to an equation is the value that makes the equation true. Once we have the value that we think is the solution, it is important to check if it is true. Let's look at a few different strategies to check solutions.
We have used  substitution to evaluate algebraic expressions. We can also use substitution to find or verify solutions to equations.
We can substitute potential solutions into an equation by replacing the variable in the equation with a number that we think could be the solution. After substituting the potential solution, we can evaluate both sides of the equation to check if the value we substituted makes the number sentence true.
If both sides of the equation equal the same value, then we know the number we substituted is a solution. If not, we need to try a different number.
By substituting the proposed solution into the equation, identify whether the following statements are true or false.
x=48 is a solution for the equation x-30=21.
x=32 is a solution to the equation x-18=14
We can use substitution to determine whether a particular value is a solution to an equation. If both sides of the equation are equal then the value is a solution.
Before we test any substitutions, we should first look at how our guess will affect the equation.
Consider the following problem: What is the solution to the equation 4y=68? Let's try y=10.
Substituting this into the left-hand side of the equation, we get:
\displaystyle \text{Left-hand side} | \displaystyle = | \displaystyle 4y |
\displaystyle \text{Left-hand side} | \displaystyle = | \displaystyle 4\times10 |
\displaystyle \text{Left-hand side} | \displaystyle = | \displaystyle 40 |
\displaystyle 40 | \displaystyle \neq | \displaystyle 68 |
Comparing this to the right-hand side of the equation, 68, we can see that our value for y needs to be a larger number.
The solution to the equation will be a value for y that is greater than 10. Let's try y=20 next.
Substituting this into the left-hand side of the equation, we have:
\displaystyle \text{Left-hand side} | \displaystyle = | \displaystyle 4y |
\displaystyle \text{Left-hand side} | \displaystyle = | \displaystyle 4\times20 |
\displaystyle \text{Left-hand side} | \displaystyle = | \displaystyle 80 |
\displaystyle 80 | \displaystyle \neq | \displaystyle 68 |
Comparing this to the right-hand side value of 68, we can see that we now need a smaller value for y. This tells us that the solution to the equation must be less than 20.
From our guesses, we know that the solution will be between 10 and 20.
Notice that, for our two guesses y=10 and y=20, the left-hand side values were 40 and 80. Another way to see that the solution to the equation is between 10 and 20 is by noticing that 68 lies between 40 and 80, and as 68 is closer to 80, our y-value will likely be closer to 20 than 10.
Consider the equation 56-t=39.
Isabelle guesses that t=10 is a solution to this equation. Is she correct?
When substituting t=10, which side of the equation is bigger?
How can Isabelle improve her guess for the solution to the equation?
Isabelle increases her guess to t=20. When substituting this into the equation she finds that 56-t is now smaller than 39. What does this tell her about the solution to the equation?
After finding a small enough range of values for the solution, we then want to test each value in the range to find the solution. We can do this by substituting each value into the equation until the equation is true.
However, this is still a fair bit of work. To save some effort, we can instead use a table of values.
Consider the equation from above: 4y=68.
As 68 is closer to 80 than 40, let's first test y=15, and if 4y is still less than 68 we know the range must be between 15 and 20. That is, we have further refined our range. We can see that 4\times15=60.
y | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 |
---|---|---|---|---|---|---|---|---|---|---|
4y | 40 | 60 |
We have now shrunk the range of values we have to test from 16 to 19.
y | 16 | 17 | 18 | 19 |
---|---|---|---|---|
4y | 64 | 68 | 72 | 78 |
We know that the solution to the equation will make the left-hand side equal in value to the right-hand side when substituted into the equation. Using the table of values, we can see that the left-hand side will be equal to 68 when y=17.
Therefore, y=17 is the solution to the equation 4y=68.
Consider the equation t+19=35.
What are the values for the left-hand side and right-hand side of the equation if Danielle substitutes in t=20?
What are the values for the left-hand side and right-hand side of the equation if Danielle substitutes in t=15?
Since Danielle knows that the solution is between 15 and 20, she decides to find the solution using a table of values. Complete the table.
t | 16 | 17 | 18 | 19 |
---|---|---|---|---|
t+19 |
Using the table of values from part (c), what value of t will make the equation t+19=35 true?
We can use a table of values to find the solution of an equation if the range of possible solutions is small enough.