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Algebra 1

Algebra 1. Chapter 2 Section 2. 2-2: Solving Two-Step Equations. Many equations have more than one operation, such as 2x + 3 = 11. Steps to solve two-step equations: 1) + or – 2) · or ÷

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Algebra 1

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  1. Algebra 1 Chapter 2 Section 2

  2. 2-2: Solving Two-Step Equations Many equations have more than one operation, such as 2x + 3 = 11. Steps to solve two-step equations: 1) + or – 2) · or ÷ Whenever you perform inverse operations, you create an equation that is equivalent to the original equation. Equivalent equations have the same solutions.

  3. Example 1: Solving 2-step Equations A) 2x + 5 = 11 - 5 - 5 2x = 6 2 2 x = 3 This equation has more than one operation. Again, look at the side where the variable is. The first step is to add or subtract. In this problem there is adding, so we do the inverse which is subtract, and subtract 5 from both sides. The 5s cancel. 11 – 5 is 6, so now, we have 2x = 6. 2x is multiplication, so we do the inverse of that, which is divide by 2 on both sides. Now, x = 3.

  4. B) 10 = 6 – 2x -6 -6 4 = - 2x - 2 - 2 - 2 = x x = - 2 Look at the side where the variable is. You see subtraction, but what is important is the sign on the 6. It is positive, so we subtract by 6 on both sides. The 6s cancel. 10 – 6 = 4, so now we have 4 = - 2x. That is multiplication, the inverse is division, so we divide by -2 on both sides. 4 divided by -2 is -2. -2 = x So, x = - 2

  5. Now you try on the white board. 1) -4 + 7x = 3 2) 1.5 = 1.2y – 5.7 3) n + 2 = 3 7

  6. Example 2: Solving Two-Step Equations That Contain Fractions Look at the side with the variable. Remember add or subtract first. We see subtraction. Even though it is a fraction, because it has the same denominator as the other side of the equation, it is easy to do. Add 1/5 to both sides. The 1/5s cancel on the left, and 3/5 + 1/5 is 4/5 on the right. Now, q/15 is division, so we multiply by 15 on both sides. The 15s cancel on the left, and the 5 cancels into 15 3 times, and 3 times 4 is 12. q = 12. Solve each equation. A) q – 1 = 3 15 5 5 + 1 +1 5 5 q = 4 15 5 q = 12 · 15 15·

  7. B) m + 2 = 9 4 - 2 - 2 m = 7 4 m = 28 Look at the side with the variable. Add or subtract first. This equation has addition, so we subtract 2 from both sides. The 2s cancel, and m/4 is division, so we multiply by 4 on both sides. The 4s cancel on the left and m = 28. · 4 4 ·

  8. Now you try on the white board. 4) 2x – 1 = 2 3 5 5 5) 3 m + 1 = 7 4 2 8 6) 1 n – 1 = 8 5 3 3

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