Substitution Method

Show your work for each problem.

1.
In the system $\begin{cases} y = 2x + 3 \\ x + y = 9 \end{cases}$ , what expression should you substitute for $y$ in the second equation?
1. a) $x + 3$
2. b) $2x - 3$
3. c) $2x + 3$
4. d) $9 - x$
Answer: $2x + 3$
The first equation tells us $y = 2x + 3$ , so we substitute $(2x + 3)$ for $y$ in the second equation.
2.
If $y = 5$ and $x + y = 12$ , what is $x$ ?
Answer: 7
Substituting $y = 5$ into $x + y = 12$ gives $x + 5 = 12$ , so $x = 7$ .
3.
Which variable is easiest to isolate in the equation $x + 3y = 15$ ?
1. a)Both are equally easy
2. b) $y$ (coefficient is 3)
3. c) $x$ (coefficient is 1)
4. d)Neither can be isolated
Answer: $x$ (coefficient is 1)
$x$ has coefficient 1, so isolating gives $x = 15 - 3y$ directly. Isolating $y$ would require dividing by 3.
4.
Solve the system: $\begin{cases} y = x + 2 \\ x + y = 10 \end{cases}$
Answer: (4, 6)
1. Substitute $(x + 2)$ for $y$ in the second equation. What equation do you get? x + x + 2 = 10
2. Simplify and solve for $x$ . What is $x$ ? 4
3. Find $y$ using $y = x + 2$ . What is $y$ ? 6
5.
Solve for $x$ : $\begin{cases} y = 3x \\ 2x + y = 15 \end{cases}$
Answer: 3
Substituting gives $2x + 3x = 15$ , so $5x = 15$ and $x = 3$ .
6.
Solve the system: $\begin{cases} x + y = 8 \\ 2x - y = 7 \end{cases}$
Answer: (5, 3)
1. From equation 1, isolate $y$ . What is $y$ equal to? 8 - x
2. Substitute into equation 2: $2x - (8 - x) = 7$ . After removing the parentheses and combining like terms you get $\square x - 8 = 7$ . What number goes in the box? 3
3. Solve $3x - 8 = 7$ for $x$ . What is $x$ ? 5
4. Find $y$ using $y = 8 - x$ . What is $y$ ? 3
7.
A student substituted $y = 2x + 1$ into $3x + y = 10$ and wrote: " $3x + 2x + 1 = 10$ , so $5x = 10$ , $x = 2$ ". What error was made?
1. a)The 1 was not subtracted from both sides: $5x + 1 = 10$ gives $5x = 9$ , so $x = 1.8$
2. b)Wrong substitution: $y$ should have been replaced by $2x - 1$
3. c)The student should have isolated $x$ first
4. d)No error: the solution is correct
Answer: The 1 was not subtracted from both sides: $5x + 1 = 10$ gives $5x = 9$ , so $x = 1.8$
The equation $5x + 1 = 10$ was incorrectly simplified to $5x = 10$ . Correct: $5x = 9$ , so $x = 1.8$ .
8.
Solve for $y$ : $\begin{cases} x = 2y - 1 \\ 3x + y = 11 \end{cases}$
Answer: 2
Substituting: $3(2y - 1) + y = 11$ , so $6y - 3 + y = 11$ , $7y = 14$ , $y = 2$ .
9.
Solve the system: $\begin{cases} 2x + y = 11 \\ x - y = 1 \end{cases}$
Answer: (4, 3)
1. From equation 2, isolate $x$ . What is $x$ equal to? y + 1
2. Substitute into equation 1: $2(y + 1) + y = 11$ . After removing the parentheses and combining like terms you get $\square y + 2 = 11$ . What number goes in the box? 3
3. Solve for $y$ . What is $y$ ? 3
4. Find $x$ using $x = y + 1$ . What is $x$ ? 4
10.
Solve for $x$ : $\begin{cases} y = 4 - x \\ 3x + 2y = 10 \end{cases}$
Answer: 2
Substituting: $3x + 2(4 - x) = 10$ , so $3x + 8 - 2x = 10$ , $x + 8 = 10$ , $x = 2$ .
11.
When you substitute and get $0 = 0$ , what does this mean about the system?
1. a)No solution (parallel lines)
2. b)Exactly one solution
3. c)Infinitely many solutions (same line)
4. d)The calculation has an error
Answer: Infinitely many solutions (same line)
When substitution gives $0 = 0$ , the equations represent the same line, so there are infinitely many solutions (any point on the line).
12.
Solve the system: $\begin{cases} 3x + 2y = 12 \\ x - y = 1 \end{cases}$
Answer: (14/5, 9/5)
1. From equation 2, isolate $x$ . What is $x$ ? y + 1
2. Substitute into equation 1: $3(y + 1) + 2y = 12$ . After removing the parentheses and combining like terms you get $\square y + 3 = 12$ . What number goes in the box? 5
3. Solve $5y + 3 = 12$ . What is $y$ ? 9/5
4. Find $x$ using $x = y + 1$ . What is $x$ ? 14/5
13.
If $\begin{cases} y = 2x - 5 \\ 4x + 3y = 15 \end{cases}$ , what is the value of $x + y$ ?
Answer: 4
Substituting: $4x + 3(2x - 5) = 15$ , so $4x + 6x - 15 = 15$ , $10x = 30$ , $x = 3$ . Then $y = 2(3) - 5 = 1$ . So $x + y = 3 + 1 = 4$ .
14.
Solve the system: $\begin{cases} 2x - 3y = 5 \\ x + 2y = -1 \end{cases}$
Answer: (1, -1)
1. From equation 2, isolate $x$ . What is $x$ equal to? -1 - 2y
2. Substitute into equation 1: $2(-1 - 2y) - 3y = 5$ . After removing the parentheses and combining like terms you get $-2 - \square y = 5$ . What number goes in the box? 7
3. Solve for $y$ . What is $y$ ? -1
4. Find $x$ using $x = -1 - 2y$ . What is $x$ ? 1
15.
A store sold 60 items for a total of 800 euros: T-shirts at 10 euros each and jeans at 20 euros each. Let $t$ be the number of T-shirts and $j$ the number of jeans. Which system describes this situation?
1. a) $\begin{cases} t - j = 60 \\ 10t - 20j = 800 \end{cases}$
2. b) $\begin{cases} t + j = 60 \\ 20t + 10j = 800 \end{cases}$
3. c) $\begin{cases} t + j = 800 \\ 10t + 20j = 60 \end{cases}$
4. d) $\begin{cases} t + j = 60 \\ 10t + 20j = 800 \end{cases}$
Answer: $\begin{cases} t + j = 60 \\ 10t + 20j = 800 \end{cases}$
$t + j = 60$ (total items) and $10t + 20j = 800$ (total revenue, price times quantity for each).
16.
Solve: A parking lot has cars and motorcycles. There are 50 vehicles and 160 wheels total. Cars have 4 wheels, motorcycles have 2. How many of each?
Answer: (30, 20)
1. Let $c$ = cars, $m$ = motorcycles. Write the equation for total vehicles. c + m = 50
2. Write the equation for total wheels. 4c + 2m = 160
3. From equation 1, isolate $m$ . What is $m$ ? 50 - c
4. Substitute and solve for $c$ . How many cars? 30
5. How many motorcycles? 20