Word problems involving systems of linear equations in two variables. Solving simultaneous equation problems by elimination and substitution. Application of simultaneous equations. Disclaimer: Some parts of the presentation are obtained from various sources. Credit to the rightful owners.

1. Applications of
Systems of Linear Equations

2. Mathematical Problems
Let 𝑥 be the larger number
𝑦 be the smaller number
𝑥 = 3𝑦
𝑥 − 𝑦 = 10
1
2
Substitution
Substitute 1 in 2
𝑥 − 𝑦 = 10
3𝑦− 𝑦 = 10
2𝑦 = 10
𝑦 = 5
Substitute 𝑦 = 5 1
𝑥 = 3𝑦
𝑥 = 3 5
𝑥 = 15
Therefore, the larger number is 15 and the smaller number is 5
1) A number is three times another number. The difference in these numbers is 10.
Find these numbers
in

3. Mathematical Problems
Let 𝑥 be the larger number
𝑦 be the smaller number
𝑥 =
1
2
𝑦
2) A number is more than half of another number by 4. The difference in these
numbers is 2. Find these numbers.
+ 4 or 𝑥 −
1
2
𝑦 = 4 1
𝑥 − 𝑦 = 2 2
Elimination
1 × 2, 2𝑥 − 𝑦 = 8 3
3 − 2 2𝑥 − 𝑦 − 𝑥 + 𝑦 = 8 − 2
= 6
Substitute 𝑥 = 6 2in
𝑥 − 𝑦 = 2
6 − 𝑦 = 2
𝑦 = 4
Therefore, the larger number is 6 and the
smaller number is 4
𝑥

4. Mathematical Problems
Let 𝑥 be the larger number
𝑦 be the smaller number
𝑦 =
1
2
𝑥
2) A number is more than half of another number by 4. The difference in these
numbers is 2. Find these numbers.
+ 4 or 𝑦 −
1
2
𝑥 = 4 1
𝑥 − 𝑦 = 2 2
Elimination
1 × 2, 2𝑦 − 𝑥 = 8 3
3 + 2 2𝑦 − 𝑥 + 𝑥 − 𝑦 = 8 + 2
= 10
Substitute 𝑦 = 10 2in
𝑥 − 𝑦 = 2
𝑥 − 10 = 2
𝑥 = 12
Therefore, the larger number is 12 and the
smaller number is 10
𝑦

5. Mathematical Problems
Let 𝑥 be the number of members in the first group
𝑦 be the number of members in the second group
𝑥 + 𝑦 = 27 1
𝑥 = 3 + 2𝑦 2
Substitution
3. To do mathematics projects, a teacher divides 27 students into two groups. One
group has 3 more than twice the number of students in the other group. Find
the number of students in both groups.
Substitute 2 in 1
𝑥 + 𝑦 = 27
3 + 2𝑦 + 𝑦 = 27
3𝑦 = 24
3𝑦 = 8
Substitute 𝑦 = 8 2in
𝑥 = 3 + 2𝑦
𝑥 = 3 + 2 8
𝑥 = 19
Therefore, the first group has 19 members while the second group has 8
members.

6. Mathematical Problems
Let 𝑥 be the length of the base of the isosceles triangle
𝑦 be the length of a leg of the isosceles triangle
𝑥 + 2𝑦 = 32 1
𝑥 = 2 +
1
2
𝑦 2
Substitution
Substitute 2 in 1
𝑥 + 2𝑦 = 32
2 +
1
2
𝑦 + 2𝑦 = 32
5
2
𝑦 = 30
3𝑦 = 12
Substitute 𝑦 = 12 2in
𝑥 = 2 +
1
2
𝑦
𝑥 = 2 +
1
2
12
𝑥 = 8
Therefore, the length of the base is 8 inches and the length of one leg is 12
inches.
4. The perimeter of an isosceles triangle is 32 inches. If the base is longer than half
of one of the other two equal sides by 2 inches, find the lengths of all sides of
this triangle.

7. Mathematical Problems
Let 𝑥 be the width of the rectangle
𝑦 be the length of the rectangle
2𝑥 + 2𝑦 = 10
2𝑥 =
1
2
𝑦
5. The perimeter of a rectangle is 10 feet. If twice the width is equal to half of the
length, find the dimensions of this triangle.
or 𝑥 =
1
4
𝑦
1
2
Substitution
Substitute 2 in 1
2𝑥 + 2𝑦 = 10
2
1
4
𝑦 + 2𝑦 = 10
5
2
𝑦 = 10
3𝑦 = 4
Substitute 𝑦 = 4 2in
𝑥 =
1
4
𝑦
𝑥 =
1
4
4
𝑥 = 1
Therefore, the width of the rectangle is 1 foot and the length is 4 feet.

8. Mathematical Problems
Let 𝑥 be the digit in the tens place
𝑦 be the digit in the ones place
10𝑥 + 𝑦 be the two-digit number
6. The digit in the ones place in a two-digit number is equal to three times the digit
in the tens place. When their places are swapped, the resulting two-digit
number is more than the original number by 54. Find the original number.
𝑦 = 3𝑥
10𝑦 + 𝑥 = 10𝑥 + 𝑦 + 54 or 9𝑦 − 9𝑥 = 54
1
2
Substitution
Substitute 1 in 2
9𝑦 − 9𝑥 = 54
9 3𝑥 − 9𝑥 = 54
18𝑥 = 54
18𝑥 = 3
Substitute 𝑥 = 3 1in
𝑦 = 3𝑥
𝑦 = 3 3
𝑦 = 9
Therefore, the original number is 39.

9. Mathematical Problems
Let 𝑥 be the digit in the tens place
𝑦 be the digit in the ones place
10𝑥 + 𝑦 be the two-digit number
10𝑥 + 𝑦 = 𝑥 + 𝑦 + 9 or 𝑥 = 1
10𝑦 + 𝑥 = 4 10𝑥 + 𝑦 − 3
7. A two-digit number is more than the sum of each digit by 9. When the digits are
swapped, the resulting number is four times less than the original number by 3.
Find this original number.
or 6𝑦 − 39𝑥 = −3
1
2
Substitution
Substitute 1 in 2
6𝑦 − 39𝑥 = −3
6𝑦 − 39 1 = −3
6𝑦 = 36
𝑦 = 6
Therefore, the original number is 16.