**Introduction**

The method of solving “by substitution” works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, “substituting” for the chosen variable and solving for the other. Then you back-solve for the first variable.

Solve the following system by substitution.

2*x* – 3*y* = –2

4*x* + *y* = 24

The idea here is to solve one of the equations for one of the variables, and plug this into the other equation. It does not matter which equation or which variable you pick. There is no right or wrong choice; the answer will be the same, regardless. But — some choices may be better than others.

For instance, in this case, can you see that it would probably be simplest to solve the second equation for “*y* =”, since there is already a *y* floating around loose in the middle there? I could solve the first equation for either variable, but I’d get fractions, and solving the second equation for *x* would also give me fractions. It wouldn’t be “wrong” to make a different choice, but it would probably be more difficult. I’ll solve the second equation for *y*:

4*x* + *y* = 24

*y* = –4*x* + 24

Now I’ll plug this in (“substitute it”) for “*y*” in the first equation, and solve for *x*:

2*x* – 3(–4*x* + 24) = –2

2*x* + 12*x* – 72 = –2

14*x* = 70

*x* = 5 Copyright © 2003-2011 All Rights Reserved

Now I can plug this *x*-value back into either equation, and solve for *y*. But since I already have an expression for “*y* =”, it will be simplest to just plug into this:

*y* = –4(5) + 24 = –20 + 24 = 4

Then the solution is (*x*, *y*) = (5, 4).

Warning: If I had substituted my “–4*x* + 24″ expression into the same equation as I’d used to solve for “*y* =”, I would have gotten a true, which appear thus,

4*x* + (–4*x* + 24) = 24

4*x* – 4*x* + 24 = 24

24 = 24

When using substitution, make sure you substitute into the *other* equation, or you’ll just be wasting your time.

Solve the following system by substitution.

*y* = 36 – 9*x*

3*x* + * ^{y}*/

_{3}= 12

We already know that these equations are actually both the same line; that is, this is a dependent system. We know what this looks like graphically: we get two identical line equations, and a graph with just one line displayed. But what does this look like algebraically?

The first equation is already solved for *y*, so I’ll substitute that into the second equation:

3*x* + (36 – 9*x*)/3 = 12

3*x* + 12 – 3*x* = 12

12 = 12

Remember that, when you’re trying to solve a system, you’re trying to use the second equation to narrow down the choices of points on the first equation. You’re trying to find the one single point that works in both equations. But in a dependent system, the “second” equation is really just another copy of the first equation, and *all* the points on the one line will work in the other line.

In other words, I got an unhelpful result because the second line equation didn’t tell me anything new. This tells me that the system is actually dependent, and that the solution is the whole line:

Solution: *y* = 36 – 9*x*

This is always true, by the way. When you try to solve a system and you get a statement like “12 = 12” or “0 = 0” — something that’s true, but unhelpful— then you have a dependent system. We already knew, that this system was dependent, but now you know what the algebra looks like.

**Solve the following system by substitution**

7*x* + 2*y* = 16

–21*x* – 6*y* = 24

Neither of these equations is particularly easier than the other for solving. I’ll get fractions, no matter which equation and which variable I choose. I’ll take the first equation, and I’ll solve it for *y*, because at least the 2 (from the “2*y*“) will divide evenly into the 16.

7*x* + 2*y* = 16

2*y* = –7*x* + 16

*y* = –( ^{7}/_{2 })*x* + 8

Now I’ll plug this into the other equation:

–21*x* – 6(–( ^{7}/_{2 })*x* + 8) = 24

–21*x* + 21*x* – 48 = 24

–48 = 24

In this case, I got a nonsense result. My entire math was right, but I got an obviously wrong answer. So what happened?

Solution: no solution (inconsistent system)

This is always true, by the way. When you get a nonsense result, this is the algebraic indication that the system of equations is inconsistent.

Note that this is quite different from the previous example. Warning: A true-but-useless result (like “12 = 12”) is quite different from a nonsense “garbage” result (like “–48 = 24”), just as two identical lines are quite different from two parallel lines. Don’t confuse the two. A useless result means a dependent system which has a solution (the whole line); a nonsense result means an inconsistent system which has no solution of any kind.

**Word Problems on Simultaneous Equation**

We have already learnt the steps of forming simultaneous equations from mathematical problems and different methods of solving simultaneous equations. In connection with any problem, when we have to find the values of two unknown quantities, we assume the two unknown quantities as x, y or any two of other algebraic symbols. Then we form the equation according to the given condition or conditions and solve the two simultaneous equations to find the values of the two unknown quantities. Thus, we can work out the problem.

**1.** The sum of two number is 14 and their difference is 2. Find the numbers.

**Solution:**

Let the two numbers be x and y.

x + y = 14………. (i)

x – y = 2………. (ii)

Adding equation (i) and (ii), we get 2x = 16

or, 2x/2 = 16/2or, x = 16/2

or, x = 8

Substituting the value x in equation (i), we get

8 + y = 14

or, 8 – 8 + y = 14 – 8

or, y = 14 – 8

or, y = 6

Therefore, x = 8 and y = 6

Hence, the two numbers are 6 and 8.

- In a two digit number. The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number.
**Solution:**Let the digit in the units place be x and the digit in the tens place be y.

Then x = 3y and the number = 10y + x

The number obtained by reversing the digits is 10x + y.

If 36 is added to the number, digits interchange their places,

Therefore, we have 10y + x + 36 = 10x + y

or, 10y – y + x + 36 = 10x + y – y

or, 9y + x – 10x + 36 = 10x – 10x

or, 9y – 9x + 36 = 0 or, 9x – 9y = 36

or, 9(x – y) = 36

or, 9(x – y)/9 = 36/9

or, x – y = 4 ………. (i)

Substituting the value of x = 3y in equation (i), we get

3y – y = 4

or, 2y = 4

or, y = 4/2

or, y = 2

Substituting the value of y = 2 in equation (i),we get

x – 2 = 4

or, x = 4 + 2

or, x = 6

Therefore, the number becomes 26.

**3.** If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions.

**Solution:**

Let the fraction be x/y.

If 2 is added to the numerator and denominator fraction becomes 9/10 so, we have

(x + 2)/(y + 2) = 9/10

or, 10(x + 2) = 9(y + 2)

or, 10x + 20 = 9y + 18

or, 10x – 9y + 20 = 9y – 9y + 18

or, 10x – 9x + 20 – 20 = 18 – 20

or, 10x – 9y = -2………. (i)

If 3 is subtracted from numerator and denominator the fraction becomes 4/5 so, we have

(x – 3)/(y – 3) = 4/5

or, 5(x – 3) = 4(y – 3)

or, 5x – 15 = 4y – 12

or, 5x – 4y – 15 = 4y – 4y – 12

or, 5x – 4y – 15 + 15 = – 12 + 15

or, 5x – 4y = 3………. (ii)

So, we have 10x – 9y = –2………. (iii)

and 5x – 4y = 3………. (iv)

Multiplying both sided of equation (iv) by 2, we get

10x – 8y = 6………. (v)

Now, solving equation (iii) and (v) , we get

10x – 9y = -2

10x – 8y = 6

– y = – 8

y = 8

Substituting the value of y in equation (iv)

5x – 4 × (8) = 3

5x – 32 = 3

5x – 32 + 32 = 3 + 32

5x = 35

x = 35/5

x = 7

Therefore, fraction becomes 7/8.

**4.** If twice the age of son is added to age of father, the sum is 56. But if twice the age of the father is added to the age of son, the sum is 82. Find the ages of father and son.

**Solution: **

Let father’s age be x years

Son’s ages = y years

Then 2y + x = 56 …………… (i)

And 2x + y = 82 …………… (ii)

Multiplying equation (i) by 2, (2y + x = 56 …………… × 2) we get

or, 3y/3 = 30/3

or, y = 30/3

or, y = 10 (solution (ii) and (iii) by subtraction)

Substituting the value of y in equation (i), we get;

2 × 10 + x = 56

or, 20 + x = 56

or, 20 – 20 + x = 56 – 20

or, x = 56 – 20

x = 36

**EXERCISES**

Lets see how much you’ve learnt, attach the following answers to the comment below

- Think of two numbers when added together give 5 and when subtracted give 1. What are the numbers? A. y = -5 and x = 10 B. y = 2 and x = 3 C. y = 5 and x = 3 D. y = 5 and x =7
- Two pens and one eraser cost Rs. 35 and 3 pencil and four erasers cost Rs. 65. Find the cost of pencil and eraser separately. A. x = 15, y = 5 B. x = 12, y = 7 C. x = 14, y = 6 D. x = 25, y = 6
- Four Pencils and six books cost ₦160. Three pencils and two books cost ₦70. Find the cost of (i) one pencil (ii) one book. A. x = 12, y = 20 B. x = 10, y = 20 C. x = 20, y = 12 D. x = 15, y = 12
- Solve the simultaneous linear equations using substitution method: 3x + y = 1 and 2x – 3y = 8 A. x = 2, y = -2 B. x = 3, y = -4 C. x = 1, y = -2 D. x = 2, y = 4
- Solve the equations x – 4y = -6 and 2x + 3y = 10 simultaneously. A. x = 2, y = 4 B. x = 2, y = 2 C. x = 3, y = 4 D. x = 4, y = 2

## 1 thought on “Classwork Series and Exercises {Mathematics- SS2}: Linear and Quadratic Solution by Substitution Method”

AnonymousX2+Y2=40 X+Y=10?