Lesson

Whenever we have an equation, we want to find the solution. The solution to an equation is the value that makes the equation true.

We have used substitution to evaluate algebraic expressions. We can also use substitution to find or verify solutions to equations.

Remember!

Substitution involves replacing one expression with another that is known to be equal to it.

We can substitute potential solutions into an equation by replacing the variable in the equation with a number that we believe may be the solution. After substituting the potential solution, we can evaluate both sides of the equation to check if the value we substituted is actually a solution. If both sides of the equation evaluate to the same value, then we know the number we substituted is a solution. If not, we need to try a different number.

Determine if the substitution $x=9$`x`=9 is a solution for the equation $\frac{63}{x}=7$63`x`=7.

**Think**: We can check if the substitution is a solution by replacing the variable $x$`x` in the equation with the numeric value $9$9. If this equation is true, $x=9$`x`=9 is a solution for the equation.

**Do**: We can substitute in our proposed solution by replacing $x$`x` in the equation with $9$9. This gives us:

$\frac{63}{9}=7$639=7

Since the quotient $\frac{63}{9}$639 can be evaluated to get $7$7, the value of the left-hand side of the equation is equal to the value of the right-hand side.

Therefore, this equation is true and the substitution $x=9$`x`=9 is a solution to the equation.

**Reflect**: After substituting the value into the equation, we compared the value of each side of the equation to see if the substitution made the equation true or not.

By substituting the proposed solution into the equation, identify whether the following statements are true or false.

$x=48$

`x`=48 is a solution for the equation $x-30=21$`x`−30=21.True

AFalse

BTrue

AFalse

B$x=32$

`x`=32 is a solution for the equation $x-18=14$`x`−18=14.True

AFalse

BTrue

AFalse

B

Before we test any substitutions, however, we should first look at how our guess will affect the expression that we are substituting it into.

Consider the expression $34+y$34+`y`.

We can substitute some values into this expression to understand how our guesses will affect it.

$y$y |
$1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|

$34+y$34+y |
$35$35 | $36$36 | $37$37 | $38$38 | $39$39 |

We can see from the table that, as the value for $y$`y` increases, the value of the expression also increases. This is because $y$`y` is being added to $34$34. This means that if we want a greater value for the expression, we should guess a greater number.

What about the expression $34-y$34−`y`?

Substituting values into this expression, we get:

$y$y |
$1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|

$34-y$34−y |
$33$33 | $32$32 | $31$31 | $30$30 | $29$29 |

We can see from this table that, as $y$`y` increases, the value of the expression actually decreases. This is because $y$`y` is being subtracted from $34$34. This means that if we want a greater value for the expression, we should guess a smaller number.

Using this method, we can also see that expressions like $4y$4`y` and $y-3$`y`−3 will increase when $y$`y` increases, while expressions like $\frac{60}{y}$60`y` and $10-y$10−`y` will decrease when $y$`y` increases.

If we are ever unsure about how a expression changes, we can just fill in a table of values!

The substitution that we test first will not always be correct, and in those cases we simply need to try again. But how do we know what to guess next?

We can improve our guess by trying to make the expression we are changing closer to the target value. When solving the equation below, we want to make the expression on the left-hand side equal to the target value on the right-hand side.

Consider the following problem: What is the solution to the equation $4y=68$4`y`=68?

Let's try $y=10$`y`=10.

Substituting this into the left-hand side of the equation, we get:

Left-hand side = $4\times10$4×10 = $40$40

Comparing this to the right-hand side of the equation, $68$68, we can see that our value for the expression $4y$4`y` needs to be increased.

From our understanding of expressions, we know that the expression $4y$4`y` will increase when $y$`y` increases. This tells us that the solution to the equation will be a value for $y$`y` that is greater than $10$10.

Although the first guess we made was not correct, it did tell us that the solution to the equation $4y=68$4`y`=68 will be greater than $10$10.

Knowing this, let's try $y=20$`y`=20 next.

Substituting this into the left-hand side of the equation, we now get:

Left-hand side = $4\times20$4×20 = $80$80

Comparing this to the right-hand side value of $68$68, we can see that we now need to decrease the value of our expression. This tells us that the solution to the equation must be less than $20$20.

From our guesses, we know that the solution will be a value for $y$`y` that is greater than $10$10 and less than $20$20.

This means that the solution will be between $10$10 and $20$20.

Notice that, for our two guesses $y=10$`y`=10 and $y=20$`y`=20, the left-hand side values were $40$40 and $80$80. Another way to see that the solution to the equation is between $10$10 and $20$20 is by noticing that $68$68 lies between $40$40 and $80$80, and as $68$68 is closer to $80$80, our $y$`y`-value is probably closer to $20$20 than $10$10.

Caution

Just because we have found a range does not necessarily mean that it is useful.

For example: $y=0$`y`=0 gives us a left-hand side value of $0$0 and $y=100$`y`=100 gives us a left-hand side value of $400$400. Since $68$68 lies between $0$0 and $400$400, the solution must lie between $y=0$`y`=0 and $y=100$`y`=100.

This is true, but not very helpful. In this case, we would like to test more values to find a smaller range.

Consider the equation $56-t=39$56−`t`=39.

Isabelle guesses that $t=10$

`t`=10 is a solution to this equation. Is she correct?Yes

ANo

BYes

ANo

BWhen substituting $t=10$

`t`=10, which side of the equation is bigger?$56-t$56−

`t`A$39$39

B$56-t$56−

`t`A$39$39

BHow can Isabelle improve her guess for the solution to the equation?

Guess a random number.

AGuess a number smaller than $10$10.

BGuess a number larger than $10$10.

CGuess $t=10$

`t`=10 again.DGuess a random number.

AGuess a number smaller than $10$10.

BGuess a number larger than $10$10.

CGuess $t=10$

`t`=10 again.DIsabelle increases her guess to $t=20$

`t`=20. When substituting this into the equation she finds that $56-t$56−`t`is now smaller than $39$39.What does this tell her about the solution to the equation?

The solution to the equation lies between $10$10 and $20$20.

AThe solution to the equation does not exist.

BThe solution to the equation is larger than $20$20.

CThe solution to the equation is smaller than $10$10.

DThe solution to the equation lies between $10$10 and $20$20.

AThe solution to the equation does not exist.

BThe solution to the equation is larger than $20$20.

CThe solution to the equation is smaller than $10$10.

D

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.

As $68$68 is closer to $80$80 than $40$40, let's first test $y=15$`y`=15, and if $4y$4`y` is still less than $68$68 we know the range must be between $15$15 and $20$20. That is, we have further refined our range. We can see that $4\times15=60$4×15=60.

$y$y |
$10$10 | $11$11 | $12$12 | $13$13 | $14$14 | $15$15 | $16$16 | $17$17 | $18$18 | $19$19 | $20$20 |
---|---|---|---|---|---|---|---|---|---|---|---|

$4y$4y |
$40$40 | $60$60 | $80$80 |

We have now shrunk the range of values we have to test from $16$16 to $19$19.

$y$y |
$16$16 | $17$17 | $18$18 | $19$19 |
---|---|---|---|---|

$4y$4y |
$64$64 | $68$68 | $72$72 | $78$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$68 when $y=17$`y`=17.

Therefore, $y=17$`y`=17 is the solution to the equation $4y=68$4`y`=68.

Consider the equation $t+19=35$`t`+19=35.

What are the values for the left-hand side and right-hand side of the equation if Danielle substitutes in $t=20$

`t`=20?Left-hand side = $\editable{}$ Right-hand side = $\editable{}$ What are the values for the left-hand side and right-hand side of the equation if Danielle substitutes in $t=15$

`t`=15?Left-hand side = $\editable{}$ Right-hand side = $\editable{}$ Since Danielle knows that the solution is between $15$15 and $20$20, she decides to find the solution using a table of values.

Complete the table:

$t$ `t`$16$16 $17$17 $18$18 $19$19 $t+19$ `t`+19$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Using the table of values from part (c), what value of $t$

`t`will make the equation $t+19=35$`t`+19=35 true?

Use substitution to determine whether a given number in a specified set makes an equation true.