Algebra Concepts: How to Solve Linear Equations with Two Unknown Values
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Algebra Concepts: How to Solve Linear Equations with Two Unknown Values

This article discusses How to Solve Linear Equations with Two Unknown Values.

As already discussed in the first article of this series, Algebra Concepts: How to solve an equation with one unknown value, its important to understand why are we using a specific set to steps to solve a problem. Mugging up the rules without understanding their importance leads to lack of proper understanding of concepts.

 

In the previous article, we discussed how to equations with one variable. In this article, our couple of equations will have two unknown variables and we’ll discuss how to solve them and figure out the value of both variables. 

 

First of all it’s important to understand that finding the unknown values in linear equations is just like a puzzle. A single linear equation is a minimal clue that is required to find the value of one variable. So, if there are two variables, we require two clues, or two equations.

 

Now, consider the two equations:

x + 2y = 8

x - y = 2

 

Here, we’ve two unknown values and two hints, now all we have to do is, combine these two hints and find out the values of x and y. 

 

The meaning of these equations is, that multiplication of y by 2 followed by addition to x will result in 8, and at the same time, the values of x and y are such that their difference is 2. That means, if x is 5, y will be 3 or may be x and y are 9 and 7 - all these pairs of x and y satisfy the second equation. But for our solution to be correct, it should satisfy the first equation too.

 

There are a couple of ways to solve these equations. Let’s solve these by substitution. We know that our solution should satisfy both the equations. Our motive now is to substitute one variable in terms of other - in other words expressing the value of x in terms of y. If in second equation, we take y to RHS, (using the complementary operations concepts we discussed in previous article) we get 

 

x = 2 + y   

 

Let’s substitute this value of x in first equation. 

 

x + 2y = 8 => (2 + y) + 2y = 8

 

Now we’ve a linear equation with one variable, 3y + 2 = 8

 

=> 3y = 6

=> y = 6 / 3 = 2

 

 Now, if we put this value of y in second equation, 

x - y = 2 

=> x - 2 = 2 

=> x = 2 + 2

=> x = 4

 

Thus, we have our final solution as 4, 2 as values of x, y respectively.

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Comments (5)
Ranked #10 in Mathematics

You need to edit the values of x and y in this paragraph: "The meaning of these equations is, that multiplication of y by 2 followed by addition to x will result in 8, and at the same time, the values of x and y are such that their difference is 2. That means, if x is 3, y will be 5 or may be x and y are 7 and 9 - all these pairs of x and y satisfy the second equation. But for our solution to be correct, it should satisfy the first equation too." Using your equation "x - y = 2" x must be larger than y, or 2 should be -2. So, either x = 5 and y = 3, or x = 3 and y = 1, or x = 9 and y = 7, etc.

Ranked #7 in Mathematics

Thanks so much Bill, for pointing out the typos - the order of the values was incorrect. Corrected them.

This is a good article. Voted this one! Hope you support my articles too.

Very well composed. Thank you for sharing your wisdom in such a manner to learn without being overwhelmed.

Good presentation . . .

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