Many people have trouble with Algebra for various reasons. Perhaps the teacher isn’t doing a great job explaining concepts, the textbook is not so great or the student is not doing enough practice. Algebra is quite easy once you get the hang of it.
In order to master algebra, you need to have a really good book or a few good books. There are also many resources online that can help you with topics you need extra help with. It is important to do a lot of practice immediately after learning a concept until the problems become easy for you to do.
Word problems, especially, seem to give a lot of students a hard time. Here are some tips on how to tackle them. Re-read the question as many times as you need to in order to understand exactly what you need to solve for and the best way to approach the problem. When you figure those out, assign the unknown or what you are trying to solve for to a variable such as x. Next you need to form algebraic expressions or equations using the information that is given, as well as your variable or variables. With the equations you can then solve for your variables using elimination, substitution, formulas and other solving methods in algebra.
Here are some examples of statements in algebra and how to convert them into an algebraic expression.
John is twice as old as Mary can be written as y=2x where y is John’s age and x is Mary’s age. Two consecutive integers can be written as x and x+1. Two consecutive even integers can be written as x and x+2 as long as x is an even number. If the sum of two numbers are 51 then the two numbers are x and 51-x. If x is a number then five more than twice a number is 2x+5.
You also need to be aware of units of measure. Make sure all the units match. For example, if you are given three numbers, two in inches and one in feet, you must covert the one in feet into inches before you can use all three numbers together to solve the unknown. A common word problem that involves converting units of measure is the coin problem. For example, Mary has several coins in her pocket including dimes, nickels, and quarters. She has three more quarters than nickels and twice as many nickels as dimes. Let’s say she has a total of $2.15.
The first thing to do is to figure out which type of coin she has the least amount of. In this case she has less dimes than nickels and quarters so let x be dimes. Since she has twice as many nickels as dimes, nickels are 2x. Since she has three more quarters than nickels and we know nickels are 2x, quarters are 2x+3. Notice we wrote quarters and nickels in terms of dimes so we can use just one variable. Now we need to convert the expressions into cents. To convert a dime into cents you multiply by 10. So we have 10x. To convert nickels into cents you multiply by 5. We have 5(2x). To convert quarters to cents multiply by 25. So we have 25(2x+3). To covert $2.15 to cents multiply by 100 to get 215 cents. Add these three expressions, set them equal to 215 and solve for x. You should get x= 2. This means she has 2 dimes, 4 nickels and 7 quarters.
In a nutshell, when given a word problem, re-read it until you fully understand what it implies. Assign a variable(s) to the unknown. Convert the statements into algebraic expressions or equations and use them to solve for the unknown. With plenty of practice you’ll become proficient at solving algebra word problems.
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