Algebraic word problems?

Question

Would someone please help me understand how to work these problems? The self-teaching guide I'm using does not provide enough examples and I'm going crazy trying to figure them out. Please explain in great detail how to translate and solve the problems. Thank you kindly.

Problem #1: Five times the smaller of two numbers decreased by 7 is equal to twice the larger number increased by 2. If the sum of the two numbers is 13, what are the numbers?

Problem #2: One number is 3 more than twice the other. What are the numbers if their sum is 18?

Last problem: The difference between two numbers is 10? Four times the larger decreased by 30 equals 3 times the smaller increased by 20. What are the numbers?

Thank you for you time and helpful insight.

Answer 1

Problem #1
Let's assume that x and y are the 2 numbers and that x is smaller than y, ok, so x is now the smaller of the two numbers and y is the larger number

So, 5x - 7 = 2y +2

Also, x + y =13

Now, we solve for x in the latter equation and get
x = -y + 13

Now, we put (-y + 13) in place of every x in the first equation

so 5(-y + 13) - 7 = 2y + 2

-5y + 65 - 7 = 2y + 2
-5y + 58 = 2y + 2
-5y + 56 = 2y
56 = 7y
8 = y

y = 8

And if y =8 then

x + 8 = 13

x = 5



Problem # 2

x and y are the numbers again

This time, one number is three more than twice the other

So, one (x) is 3 more than twice the other (y)

so x = 2y +3

Also, they add to equal 18

x + y =18

Solve for y in the latter equation, x would work also, but y will make it simpler here

so y = -x + 18

Now for every y we put (-x + 18) in the first equation

x = 2(-x + 18) + 3

x = -2x + 36 + 3

3x = 39

x = 13

and using x + y = 18

13 + y = 18

y = 5



Problem #3
X and y are the two numbers

Difference between them is 10
We will say X is the larger of the two

X - Y = 10

And 4*larger number(x) - 30 equals 3*smaller number(y) + 20
Basically

4x - 30 = 3y + 20

We solve the first equation for x or y, i choose x

x = y+10

Now for every x we put (y+10)

4(y+10) - 30 = 3y + 20

4y + 40 - 30 = 3y + 20
4y + 10 = 3y + 20
4y = 3y + 10

y = 10

Since y = 10, using x -y = 10

x - 10 = 10

x = 20

Hope I could Help

Answer 2

1. smaller = x, larger = y. There are two ways of reading this one. One is 5 * (x - 7) = 2 (y + 2) and the other is 5x - 7 = 2y + 2

i. x + y = 13
minus y from both sides:
x = 13 - y

5 * (x - 7) = 2 (y + 2)
Substitute x = 13 - y
5 * ((13 - y) - 7) = 2y + 4
Expand brackets
5 * (6 - y) = 2y + 4
30 - 5y = 2y + 4
Plus 5y and minus 4 from both sides
26 = 7y
y = 26/7 = 3 and 5/7ths
Substitute into x + y = 13
x = 65/7 = 9 and 2/7ths

ii. 5x - 7 = 2y + 2
Substitute in x = 13 - y
5 (13 - y) - 7 = 2y + 2
Multiply out
65 - 5y - 7 = 2y + 2
58 - 5y = 2y + 2
Plus 5y minus 2 from both sides
56 = 7y
y = 56/7 = 8
Substitute into x + y = 13
-> x = 13 - 8 = 5

2. One number = p, the other = q
p = 2q + 3
p + q = 18
Therefore p = 18 - q
Substitute into p = 2q + 3
18 - q = 2q + 3
Add q to both sides
18 = 3q + 3
Minus 3 from both sides
15 = 3q
-> q = 15/3 = 5
q = 5
Substitute into p + q = 18
q + 5 = 18
-> q = 18 - 5 = 13
q- 13

3. Larger = a, smaller = b
4a - 30 = 3b + 20
a - b = 10
Add b to both sides
a = 10 + b
Substitute into 4a - 30 = 3b + 20
4 (10 + b) - 30 = 3b + 20
Expand brackets
40 + 4b - 30 = 3b +20
10 + 4b = 3b + 20
Minus 20 from both sides
-10 + 4b = 3b
Minus 3b from both sides
-10 + b = 0
-> b = 10
Substitute into a - b = 10
a - 10 = 10
Add 10 to both sides
a = 20

Answer 3

Okay - in all of these problems you have 2 numbers. They tell you the relationship of the two numbers in one way or the other.

In the first problem, for example, it tells you the sum of the two numbers is 13. So if you had two numbers x and y then x + y = 13

However, at this stage you probably don't want to be messing around with two variables, x and y so you might want to call the two numbers x and 13-x (do you see how that happens? if you add them they make 13)

Then you need to make use of the other information

5 times the smaller one and then subtract 7 is 5(13-x) - 7
and it is equal to 2x + 2

So you make your equation.



Same sort of thing for problem 2
You could say that the "other" number could be x and then the first number is 3 more than twice the other, ie. 2x+3 and now you add them together and get 18 - this is one way.

Alternatively you could use the same technique as (1) and call the numbers x and 18-x

Answer 4

2) x = first number and y = second number

x = 3 + 2*y
x + y = 18

substitute x into the 2nd equation

(3 + 2y) + y =18
3 + 3y = 18
3y = 15
y = 5

x = 3 +2y
x = 3 + 2*5
x = 3 + 10
x = 13

the numbers are 13 and 5

3)

x - y = 10
4*x - 30 = 3*y +20

x = 10 + y
4*(10+y) - 30 = 3y + 20
40 + 4y - 30 = 3y + 20
10 + 4y = 3y + 20
4y - 3y = 20 -10
y = 10

x = 10 + y
x = 10 +10
x = 20

The numbers are 10 and 20

Answer 5

#1
5*x - 7 = 2*y + 2
x+y = 13

# 2
y = 3 + 2*x
x + y = 18

#3
x - y = 10
4*x - 30 = 3*y + 20

I'll assume you can do the algebra to finish these!

Answer 6

Problem #2

x=3 + 2y (because it says one # (x) is 3 more than twice the other (3+2y)
x + y=18 (it says they add up to be 18)
3+2y+y = 18 ( u can substitute in 3 + 2y for x because u knw they're equal
3y=15 ( u subtract 3 on both sides and add the two values of 'y')
y=5 (you divide both sides by 3 to get this)

hope this helped!!

Answer 7

Ask your maths teacher or tuition teacher tomorrow.