3 Variable System of Equations Worksheets - Math Monks - Free Printable

Let's solve each of the four problems step by step.

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Problem 1:

The sum of the digits of a three-digit number is 16. If the ten's digit of the number is 3 times the unit's digit and the unit's digit is one-fourth of the hundredth digit, then what is the number?

#### Step 1: Define variables
Let:
- $ h $ = hundreds digit
- $ t $ = tens digit
- $ u $ = units digit

We are given:
1. $ h + t + u = 16 $ → (Equation 1)
2. $ t = 3u $ → (Equation 2)
3. $ u = \frac{1}{4}h $ → (Equation 3)

#### Step 2: Substitute Equation 3 into Equation 2
From (3): $ h = 4u $

Now plug into (2): $ t = 3u $

Now substitute $ h = 4u $, $ t = 3u $ into (1):

$$
4u + 3u + u = 16 \\
8u = 16 \\
u = 2
$$

Then:
- $ t = 3u = 6 $
- $ h = 4u = 8 $

So the number is: $ 862 $

✔ Answer: 862

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Problem 2:

If the equation of a parabola is generally expressed as $ y = ax^2 + bx + c $, find the equation of one that passes through the points (-2, 40), (1, 7), (3, 15).

We will plug in each point to get a system of equations.

#### Step 1: Plug in the points

For $(-2, 40)$:
$$
40 = a(-2)^2 + b(-2) + c = 4a - 2b + c \quad \text{(Eq 1)}
$$

For $(1, 7)$:
$$
7 = a(1)^2 + b(1) + c = a + b + c \quad \text{(Eq 2)}
$$

For $(3, 15)$:
$$
15 = a(3)^2 + b(3) + c = 9a + 3b + c \quad \text{(Eq 3)}
$$

#### Step 2: Solve the system

We have:
1. $ 4a - 2b + c = 40 $
2. $ a + b + c = 7 $
3. $ 9a + 3b + c = 15 $

Let’s eliminate $ c $.

Subtract Eq 2 from Eq 1:
$$
(4a - 2b + c) - (a + b + c) = 40 - 7 \\
3a - 3b = 33 \\
a - b = 11 \quad \text{(Eq 4)}
$$

Subtract Eq 2 from Eq 3:
$$
(9a + 3b + c) - (a + b + c) = 15 - 7 \\
8a + 2b = 8 \\
4a + b = 4 \quad \text{(Eq 5)}
$$

Now solve Eqs 4 and 5:

From Eq 4: $ a = b + 11 $

Plug into Eq 5:
$$
4(b + 11) + b = 4 \\
4b + 44 + b = 4 \\
5b = -40 \\
b = -8
$$

Then $ a = -8 + 11 = 3 $

Now use Eq 2: $ a + b + c = 7 $
$$
3 - 8 + c = 7 \\
-5 + c = 7 \\
c = 12
$$

✔ So the equation is:
$$
y = 3x^2 - 8x + 12
$$

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Problem 3:

There are 12 pieces of five, ten, and twenty rupee coins whose total value is ₹105. When the number of five and ten rupee coins are interchanged, the total value increases by ₹20. Find the number of coins of each type.

Let:
- $ x $ = number of ₹5 coins
- $ y $ = number of ₹10 coins
- $ z $ = number of ₹20 coins

We are given:
1. $ x + y + z = 12 $ → (Eq 1)
2. $ 5x + 10y + 20z = 105 $ → (Eq 2)
3. After interchanging $ x $ and $ y $, total becomes ₹125:
$$
5y + 10x + 20z = 125 \quad \text{(Eq 3)}
$$

Let’s simplify these equations.

Divide Eq 2 by 5:
$$
x + 2y + 4z = 21 \quad \text{(Eq 2')}
$$

Divide Eq 3 by 5:
$$
y + 2x + 4z = 25 \quad \text{(Eq 3')}
$$

Now we have:
- (Eq 1): $ x + y + z = 12 $
- (Eq 2'): $ x + 2y + 4z = 21 $
- (Eq 3'): $ 2x + y + 4z = 25 $

Let’s subtract Eq 1 from Eq 2':
$$
(x + 2y + 4z) - (x + y + z) = 21 - 12 \\
y + 3z = 9 \quad \text{(Eq 4)}
$$

Now subtract Eq 2' from Eq 3':
$$
(2x + y + 4z) - (x + 2y + 4z) = 25 - 21 \\
x - y = 4 \quad \text{(Eq 5)}
$$

Now solve:
From Eq 5: $ x = y + 4 $

From Eq 4: $ y = 9 - 3z $

Now plug into Eq 1:
$$
x + y + z = 12 \\
(y + 4) + y + z = 12 \\
2y + z + 4 = 12 \\
2y + z = 8
$$

But $ y = 9 - 3z $, so:
$$
2(9 - 3z) + z = 8 \\
18 - 6z + z = 8 \\
18 - 5z = 8 \\
5z = 10 \\
z = 2
$$

Then:
- $ y = 9 - 3(2) = 3 $
- $ x = y + 4 = 7 $

Check:
- Total coins: $ 7 + 3 + 2 = 12 $ ✔
- Value: $ 5(7) + 10(3) + 20(2) = 35 + 30 + 40 = 105 $ ✔
- After interchange: $ 5(3) + 10(7) + 20(2) = 15 + 70 + 40 = 125 $ → increase by ₹20 ✔

✔ Answer:
- ₹5 coins: 7
- ₹10 coins: 3
- ₹20 coins: 2

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Problem 4:

Rob has a bag containing 35 nickels and some quarters. The total value of these coins is less than $2.50. What is the maximum number of quarters that meets the given condition?

Let:
- Number of quarters = $ q $

Value:
- Each nickel = $0.05 → 35 nickels = $ 35 × 0.05 = $1.75
- Each quarter = $0.25 → $ 0.25q

Total value: $ 1.75 + 0.25q < 2.50 $

Solve inequality:
$$
1.75 + 0.25q < 2.50 \\
0.25q < 0.75 \\
q < 3
$$

Since $ q $ must be an integer, maximum $ q = 2 $

✔ Answer: 2 quarters

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✔ Final Answers:

1. 862
2. $ y = 3x^2 - 8x + 12 $
3. ₹5 coins: 7, ₹10 coins: 3, ₹20 coins: 2
4. Maximum number of quarters: 2

Let me know if you'd like explanations in simpler terms or visual steps!