Substitution Methods
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.
2x – 3y = –2
4x + 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, 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? We 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. But let us solve the second equation for y:
4x + y = 24
y = –4x + 24
Now I’ll plug this in (“substitute it”) for “y” in the first equation, and solve for x:
2x – 3(–4x + 24) = –2
2x + 12x – 72 = –2
14x = 70
x = 5 Copyright © -2011 All Rights Reserved
Now I can plug this x-value back into either equation, and solve for y. But since we 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).
But if we had substituted our “–4x + 24″ expression into the same equation as we’d used to solve for “y =”, we would have gotten a true, but useless, statement:
4x + (–4x + 24) = 24
4x – 4x + 24 = 24
24 = 24
Twenty-four does equal twenty-four, but who cares? So 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 – 9x
3x + 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:
3x + (36 – 9x)/3 = 12
3x + 12 – 3x = 12
12 = 12
Yes, twelve equals twelve, but so what?
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 – 9x
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, from the previous lesson that this system was dependent, but now you know what the algebra looks like.
Solve the following system by substitution.
7x + 2y = 16
–21x – 6y = 24
Neither of these equations is particularly easier than the other for solving. We’ll get fractions, no matter which equation and which variable I choose. I guess we’ll take the first equation, and we’ll solve it for y, because at least the 2 (from the “2y“) will divide evenly into the 16.
7x + 2y = 16
2y = –7x + 16
y = –( 7/2 )x + 8
Now we’ll plug this into the other equation:
–21x – 6(–( 7/2 )x + 8) = 24
–21x + 21x – 48 = 24
–48 = 24
In this case, we got a nonsense result. The entire math was right, but we got an obviously wrong answer. So what happened?
Keep in mind that, when solving, you’re trying to find where the lines intersect. What if they don’t intersect? Then you’re going to get some kind of wrong answer when you assume that there is a solution (as I did when I tried to find that solution). We knew, from the previous lesson, that this system represents two parallel lines. But I tried, by substitution, to find the intersection point anyway. And I got a “garbage” result. Since there wasn’t any intersection point, my attempt led to utter nonsense.
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.
If two (or more) equations have the same variables and the same solutions then they are simultaneous equations. For example, these equations are simultaneous equations:
x + y = 3 and
2x + 3y = 8
because both have the same variables: ‘x’ and ‘y’, and the same solutions: x = 1, y = 2
Substituting x = 1 and y = 2 into both equations, they BOTH give correct answers:
1 + 2 = 3 and
2´1 + 3´2 = 8
Thus: x = 1 and y = 2 are the solutions to both equations. ‘Solving’ simultaneous equations means finding the values of ‘x’ and ‘y’ that make them true. The following steps will demonstrate how to solve simultaneous equations by the substitution method.
We will use the example equations above to demonstrate the procedure…
(1) Isolate one of the variables ( ‘x’ ) on one side of one of the equations:
x + y = 3
Isolating ‘x’:
x = 3 – y
(2) Substitute for the isolated variable in the other equation:
2x + 3y = 8
Substituting 3 – y for ‘x’:
2(3 – y) + 3y = 8
This equation has only one variable, so we can solve it.
(3) Solve this equation for the other variable, ‘y’:
2(3 – y) + 3y = 8
Expanding the brackets:
6 – 2y + 3y = 8
Simplifying:
6 + y = 8
Subtracting 6 from both sides:
y = 2
(4) Substitute the known value of ‘y’ into the equation for ‘x’ derived in step 1:
x = 3 – y
Substituting 2 for ‘y’:
x = 3 – 2
Therefore:
x = 1
Word Problems on Simultaneous Equation
Solving the solution of two variables of system equation that leads for the word problems on simultaneous linear equations is the ordered pair (x, y) which satisfies both the linear equations.
Problems of different problems with the help of linear simultaneous equations:
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.
Worked-out examples for the word problems on simultaneous linear equations:
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 is 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 = 2Substituting 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
5. 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.
Solution:
Let the cost of pen = x and the cost of eraser = y
Then 2x + y = 35 ……………(i)
And 3x + 4y = 65 ……………(ii)
Multiplying equation (i) by 4,
Subtracting (iii) and (ii), we get;
5x = 75
or, 5x/5 = 75/5
or, x = 75/5
or, x = 15
Substituting the value of x = 15 in equation (i) 2x + y = 35 we get;
or, 2 × 15 + y = 35
or, 30 + y = 35
or, y = 35 – 30
or, y = 5
Therefore, the cost of 1 pen is Rs. 15 and the cost of 1 eraser is Rs. 5.
