Q&A for How to Solve Two Step Algebraic Equations

The goal is to get x on one side of the equation. So add 8 to both sides of the equation; x/5 - 8 + 8 = 7 + 8; x/5 = 15. Next, multiply both sides of the equation by 5. 5 (x/5) = 15 (5). x = 75.

5x - 6 = 3x - 3. Subtract 3x from both sides: 2x - 6 = -3. Add 6 to both sides: 2x = 3, so x = 3/2.

9x² -12xy+4y². It's figured this way: (3x)(3x) = 9x². (3x)(-2y) = -6xy, and (-2y)(3x) = -6xy; (-6xy) + (-6xy) = -12xy. (-2y)(-2y) = +4y².

Multiply both sides by 4 to get rid of the fraction: x + 28 = -36. Subtract 28 from both sides: x = -64.

f = [(9/5)C] + 32.

Start with 10 - y = 2. Adding y to both sides and subtracting 2 from both sides, we see that 8 = y. Then we insert that value of y into x + y = 2 and find that x = -6. Check by inserting those values of x and y into the full equation: -6 + 8 = 10 - 8 = 2.

Let a variable (such as x) stand for the number you're looking for. Set up an equation in x that expresses the conditions described in the problem, and solve the equation for x. It takes some practice to become good at setting up such an equation, because every problem is slightly different. Ask for help from someone who knows how to do it, and then keep practicing.

That's fine. The answer does not have to have any operation.

Subtract 9 from both sides of the equation, then divide both sides by 7.

First simplify both sides of the equation: 4x + 1 = 9 - 4x. Then add 4x to both sides and subtract 1 from both sides: 8x = 8. Then divide both sides by 8.

Add 3x to both sides of the equation, and subtract 2 from both sides. That leaves you with 15 = 3x, so that x = 5.

You should practice solving them every day. You can also ask a friend, teacher, or parent for help if you need assistance when you are getting started.

(2z - 4 - z) is the same as (z - 4). So z - 4 = 4. Add 4 to both sides of the equation, making z equal to 8.

First, multiply (h + 28) by 20, which gives you 20h + 560 = 740. Next, subtract both sides by 560; this gives you 20h = 180. Divide both sides by 20, and you'll get the answer: h = 9.

Multiply both sides of the equation by 4: k - 3 = 40. Add 3 to both sides: k = 43.

Suppose there is an equation x^2-x=2. Then,x^2-x-2=0.Now,x^2-x-2 can be represented as x^2+x-2x-2. So,if you factorize this expression, you get x(x+1)-2(x+1)=0.then, (x-2)(x+1)=0.so,x=1,-2.so,in general you factorize the Lhs after converting the rhs to 0.

If the variable is a negative, simply multiply by negative 1 to make it positive.

First you want to isolate the m. Multiply both sides by 3 and you get 9+m=6. Finally, subtract both sides by nine and you get m=-3.

(a + 9) / a = 3. Multiply both sides of the equation by a: a + 9 = 3a. Subtract a from both sides: 9 = 2a. Divide both sides by 2: 9/2 = a. a³ = (9/2)(9/2)(9/2) = 729 / 8 = 91 1/8.

Multiply both sides of the equation by (289 + x). Then you'll have to use the quadratic formula.

4x + 16 = 24. Subtract 16 from both sides of the equation: 4x = 8. Divide both sides by 4: x = 2.

The easiest way is to use the quadratic formula.

If x= -2, x³ + 9x + 10 = (-2)³ + (9)(-2) + 10 = (-8) -18 +10 = -16.

2x² + 7 = 15. 2x² = 8. x² = 4. x = ± 2.

Since that expression is not an equation, you can't solve for v.

We're solving for x. -3 ≥ 4 - 3x. Subtract 4 from both sides: -7 ≥ -3x. Divide both sides by -3: 7/3 ≥ x. Therefore, x can be any number equal to or less than 7/3.

Multiply both sides of the equation by -2. Then x - 5 = 14, and x = 19.

Subtract 2/5 from both sides of the equation. Then a/5 = 2/5. By inspection (or by cross-multiplication), a = 2.

Subtract 7 from both sides. Then multiply both sides by 3. a = 18.

(4a / 7) + 5 = 0. Multiply both sides of the equation by 7: 4a + 35 = 0. Subtract 35 from both sides: 4a = -35. Divide both sides by 4: a = -35/4.