Linear Equations with 2 unknowns

by Carson
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In the article of Algebra, we’ve talked about equations with 1 unknown. But we can have 2 or more variables. Then, how to solve them? There are 2 ways. A linear equation should look like this:

\[x + y = 8\] \[y = x – 1\]

Substitution

The first way to do that is by substitution. That means one of the two variables get replaced by an expression with another unknown. If you do that, congratulations! You can figure out the value of the unknown which is not replaced, and calculate another based on the data.

In the last example, it is clear that we will use substitution. Remember to replace y with x – 1!

\[x + y = 8\] \[y = x – 1\] Then: \[x + x – 1 = 8\] \[2x – 1 = 8\] \[2x = 9\] Finally, \[x = 4.5\] Calculate \[4.5 + y = 8\] \[y = 3.5\]

This is just an example. You can apply x and y wherever you choose. But of course, in this scenario, picking y below is the most convenient choice and what you should do.

Elimination

Next, there is the method of elimination. It is like when you do addition or subtraction of fractions, you need to make the denominator the same by multiplying them with a certain number. Then, if the associated number carried by the same variable is identical, you can add or subtract a whole equation. In the same way with substitution, you can only figure out 1 variable directly, while the other is calculated by the data.

\[2x + 7y = 22\] \[8x + 5y = 42\] Then, both values associated with x should be 8x. \[8x + 28y = 88\] After that, the answer starts to appear \[23y = 46\] \[y = 2\] Figure out x by the value of y \[2x + 14 = 22\] \[2x = 8\] Finally, \[x = 4\]

Also, you can apply for all variables anywhere but this is probably the best way to do that.

The Number of Solutions

Moreover, as they have 2 or more variables, sometimes the 2 equations “argue”, making them unable to solve. You may not know when equations can “argue”, but what if you get the step of 15 = 18? Well, the 2 equations “argued” again.

If they have not argued but the 2 formulas aren’t identical, it has only 1 solution. And if the 2 formulas turn out to be identical (the numbers also needs to be identical for the variables), it has an infinite number of solution, that means it is similar to an identity.

\[x + y = 4z\] \[2x + 2y = 8z\] Then, \[2x + 2y = 8z\] Are they identical? Are the numbers associated with the variables the same? Is the one after the = mark a variable, not an integer?

In this scenario, three answers are yes. Therefore, it has an infinite number of solutions, meaning that any values for x and y are correct. How about this?

\[x + y = 5\] \[4x + 4y = 10\] After that, \[4x + 4y = 20\] Oh no! Guess what happened!

10 isn’t equal 20, so the linear equations have no solutions. You can’t solve this, you can only point out that the question has some problems unless it is asking you about this. And finally, this example should be the equations you will solve.

\[2x + 3y = 7\] \[4x + 7y = 9\] You can use either substitution or elimination, but elimination is probably the best way to solve that in this scenario. \[4x + 6y = 14\] Hence: \[y = -5\] Then, “plug in” -5 into either equation. \[x = 11\]

Real-life usage

These equations can help point out the missing number without counting. For example, if you had a farm and you forgot the number of a certain animal, but you remember the number of legs for animals and the number of animals. How to figure out this? Well, solving the associated linear equation can save your day.

\[2x + 4y = 68\] \[x + y = 20\] Then, \[2x + 2y = 40\] We are already collecting more data. \[y = 14\] Finally, plug in 14 to the equation \[x = 6\]

And also, we can learn how to solve other problems in your life based on the basic knowledge of the linear equation! You can still use elimination and substitution in linear equations with 3 or 4 unknowns, and so on.

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