Q&A for How to Learn Algebra

Basic math skills you learned in elementary or primary school are the fundamentals of algebra. This includes concepts like adding, subtracting, multiplying and dividing.

Subtract 13 from both sides to get x by itself. That makes the equation: x = 24 - 13 or: x = 11.

No, because the first equation asks for addition and the second equation asks for multiplication.

Algebra is a good tool for solving mathematical puzzles and situations that may arise in real life. Think about how you could apply it to your daily life.

Use the quadratic formula, [-b +/- √(b² - 4ac)] / 2a. To solve for x, evaluate the formula with a = 3, b = -5, and c = -1.

Isolate the variable on one side of the equation and the constant on the other side. In this example, subtract 3x from both sides, leaving no x on the left side and 3x on the right side. Then add 7 to both sides, leaving no constant on the right side and 11 on the left side. Then divide both sides by the remaining coefficient of the variable. That leaves 11/3 on the left side and x on the right side. 11/3 is the value of x. Check the answer by substituting 11/3 for each x in the original equation and seeing that each side of the equation equals the other side. In this example, (3)(11/3) + 4 = 15, and (6)(11/3) - 7 also equals 15.

Is x the exponent or the base? If it is the base, you will most likely have to factor it or use the quadratic equation. If the x is cubed, there are formulas you can memorize. If it raised to the fourth power, it often cannot be done by hand, unless by factoring. If x is the exponent, you have to use logarithms.

Multiply (b-1) by 7. Then subtract (8-b), which means you subtract 8 and add b.

Subtract 5d from both sides, and add 3 to both sides: 4d = 20. Divide both sides by 4: d = 5.

That's (3.3)(3.3)(3.3) = (10.89)(3.3) = 35.937.

Place positive integers to the right of zero, negative integers to the left of zero. For example, +17 is positioned 17 units to the right of zero, and -9 is positioned 9 units to the left of zero.

z = 4.0 = y/x = 5.6 / x. So 4x = 5.6, and x = 1.4.

Because this is not an equation, it can't be "solved." All you can do is perform the indicated multiplication: (-7)(9x - 11) = -63x + 77.

Combine both like terms, then begin solving. Depending on the problem, this will generate either a single-step, double-step, or multi-step equation. If something such as 2(x + 3) appears, distribute the 2 to both x and 3 (use multiplication) and you'll get 2x + 6. Then combine the like terms.

I assume you mean (x^8 + y^8) / (x^4 y^4)? The trick to this one is to exploit the symmetry in the expressions and try to get to the answer directly (without trying to solve for either x or y). There are other valid approaches, but this sort of problem is usually found on exams with too brisk of a time limit for raising irrational numbers to the 8th power by hand. Start by squaring your given equation to x^2/y^2 + y^2/x^2 - 2 = 9. Add 2 to combine the constants and square it again to x^4/y^4 + y^4 / x^4 + 2 = 121. So x^4 / y^4 + y^4 / x^4 = 119 = (x^8 + y^8) / x^4 y^4.

Algebra is not typically introduced in primary school. (Something often called "pre-algebra" may be introduced, but not algebra.) If by the time you leave primary school you understand most of the concepts shown in Part 1 above, you should be ready for algebra.

2x + 3 = 7. Subtract 3 from both sides: 2x = 4. Divide both sides by 2: x = 2.

Yes. Adding a negative 4 is the same thing as subtracting a (positive) 4.

If you're asking how to get rid of a radical sign in an equation, the above article does a pretty good job of explaining it. To get rid of a square root, square it. To get rid of a cube root, cube it. To get rid of a fourth root, raise it to the fourth power, and so on. For example, if the fifth root of x equals 2, you would solve for x by raising both sides of the equation to the fifth power. On one side of the equation, the fifth root of x raised to the fifth power is x. On the other side of the equation, the fifth power of 2 is 32. So x = 32.

Not really. As with any skill, it just takes practice.

You can't "solve" this because it's not an equation. This, however, can be factored into 6(a - 2).

Because that is an equation with three variables in it, all you can do is solve for x in terms of e and y (meaning that e and y will appear in the solution, and you won't have a numerical solution, since e and y do not have specific values). The solution, such as it is, involves first performing the subtraction on the right side of the equation, then dividing both sides of the equation by the two non-x factors on the left side. Then divide both sides by 7, leaving only x² on the left side. Finally, take the square root of each side, and you'll have the value of x on the right side. (The value will be "±.")

Do whatever it takes to "isolate" the variable (that is, to let it stand alone on one side of the equals sign). A simple example of solving for x: x+4 = 0. Subtract 4 from both sides: x = -4. Another example: x-5 = 2. Add 5 to both sides: x = 7.

There's really no "easy" way to do it. It's a matter of working hard until you understand the concepts. If you've been struggling, you would probably benefit from some tutoring. Ask for help from a teacher, a friend or a family member, or get a recommendation for a tutor with a demonstrated ability to explain concepts clearly.

Add 5500y to both sides of the equation, and subtract 67,000,000 from both sides. Now you have 5500y = x - 67,000,000. Then divide both sides by 5500 to get y = (x - 67,000,000) / 5500. That's as far as you can go without knowing the value of x.

Algebra is an attempt to solve problems that involve unknown quantities. It is often used in scientific, technological and industrial applications.

Because algebra is taught in high school, colleges don't usually address the subject unless it's part of a remedial course.

You might if you're really motivated.

Buy or borrow a beginning-algebra textbook. They're available on Amazon and at bookstores and libraries.

See Solve Simultaneous Equations Graphically , Solve Simultaneous Equations Using Substitution Method , Solve Simultaneous Equations Using Elimination Method and related articles.