Solving Equations Conceputally

Solving Equations Conceputally
Dr. Sarah Ledford Mathematics Educator

Solving Equations standards/8/EE/C/7/tasks/392 In elementary school, students often draw pictures of the arithmetic they do. For instance, they might draw the following picture for the problem 2 + 3: In this picture, each square represents 1 tile.

Solving Equations We can do the same thing for algebraic expressions, but we need to be careful about how we represent the unknown. If we assume that an unknown number of tiles x are contained in a bag, we could draw the following picture for 2x + 3:

Solving Equations When we have an equation to solve, we assume that the two sides of the equation are equal. We can represent this by showing them level on a balance. The equation 2x + 3 = 7 could be shown as:

Solving Equations When we solve equations, we can add, subtract, multiply or divide both sides of the equation by the same thing in order to maintain the equality. This can be shown in pictures by keeping the balance level. We could solve the equation 2x + 3 = 7 using pictures by first removing (subtracting) 3 from each side, and then splitting (dividing) the remaining blocks into two equal groups. We’ll come back to this one…

Homemade Manipulatives
Blue is positive. White is negative. Each counter represents 1 or −1. The “bag of pebbles” represent an unknown amount – we do not know how many “counters” are inside the bag. Each bag of pebbles represents x or −x.

Neutrals or Zero Pairs When working with signed numbers, we also let a “neutral” be represented by “+/− “ because it has no value and does not change the value of the problem in any way. = 0 = 0

x + 2 = 7 Solve by “inspection.” That simply means that you can solve the problem by looking at it. If you know your fact families, it shouldn’t be too difficult. Now, let’s use manipulatives. Remember that our goal is to get x by itself on one side.

= x + 2 7 What should we put here? What should we put here?

= x + 2 7

BALANCE We must keep the equation balanced! If we take away 1 from the left-hand side (LHS), we must take away 1 from the right-hand side (RHS). If we add 2 to the LHS, we must add 2 to the RHS. These are the addition & subtraction properties of equality.

BALANCE We must keep the equation balanced! If we multiply the LHS by 4, we must multiply the RHS by 4. If we divide the LHS by 2, we must divide the RHS by 2. These are the multiplication & division properties of equality.

= x + 2 7 Take away 1 from each side

= x + 2 7 x = 5

We took away 2 from each side.
x + 2 = 7 We took away 2 from each side. x + 2 = 7 – 2 – 2 x = 5 Writing it out symbolically is VERY important as that is our goal (aside from getting the correct answer)!!

Solve by “inspection.” Now, let’s use manipulatives.
x – 3 = 5 Solve by “inspection.” Now, let’s use manipulatives.

x – 3 = 5 From our work with signed numbers, we know this is the same as: x + (−3) = 5 This is a SUPER important idea!! We can represent the LHS of this with an x and 3 negatives. Then we can add 3 positives to both sides to neutralize the 3 negatives on the LHS.

= x – 3 5

= x – 3 5 Add 3 to each side

= x – 3 5 Pair neutrals or zero pairs

= x – 3 5 Clear away neutrals or zero pairs

= x – 3 5 x = 8

x – 3 = 5 We added 3 to each side. x – 3 = 5 x = 8

2x = 8 Solve by “inspection.” Now, let’s use manipulatives.

= 2x 8 This means that there are 8 in 2 groups.
We want to know how many are in 1 group. We need to divide 8 into 2 equal groups.

Do you see the 2 equal groups of 4?
= 2x 8 Do you see the 2 equal groups of 4?

= 2x 8 x = 4

We divided both sides by 2.
2x = 8 We divided both sides by 2. 2x = 8

2x = 8 We divided both sides by 2. 2x = x = 4

3x = −6 Solve by “inspection.” Now, let’s use manipulatives.

= 3x −6 This means that there are −6 in 3 groups.
We want to know how many are in 1 group. We need to divide −6 into 3 equal groups.

Do you see the 3 equal groups of -2?
= 3x −6 Do you see the 3 equal groups of -2?

= 3x −6 x = -2

3x = −6 We divided both sides by 3. 3x = −6

3x = −6 We divided both sides by 3. 3x = − x = −2

3x + 2 = 5 How is this equation different than what we have been doing? The previous were one-step equations. This is a multi-step equation that requires more than one- step to solve it. These are not as easy to solve by inspection. Let’s use manipulatives.

= 3x + 2 5 What do you think we should do first?
What is common on both sides? Take away 2 blue chips

Now we need to divide both sides into 3 equal groups.
= 3x + 2 5 Now we need to divide both sides into 3 equal groups.

= 3x + 2 5

= 3x + 2 5 x = 1

3x + 2 = 5 We took away 2 from each side. 3x + 2 = 5 – 2 – 2 3 x = 3 Then we divided what was left into 3 equal groups. 3 x = 3

3x + 2 = 5 We took away 2 from each side. 3x + 2 = 5 – 2 – 2 3 x = 3 Then we divided what was left into 3 equal groups. 3 x = x = 1

2(x + 1) = 10 1 + 1 = 2 2 ones x + x = 2x 2 x’s 2(x + 1) = two (x + 1)’s 2(x + 1) = (x + 1) + (x + 1) 2(x + 1) = x x + 1 2(x + 1) = 2x + 2

= 2(x + 1) 10 NOTE: two (x + 1)’s is the same as 2x + 2.
Distributive Property!! Take 2 blue chips from each side.

We need to divide what is left into 2 equal groups.
= 2(x + 1) 10 We need to divide what is left into 2 equal groups.

= 2(x + 1) 10

= 2(x + 1) 10 x = 4

We saw that 2(x + 1) is the same as 2x + 2.
– 2 – 2 2x = 8

We saw that 2(x + 1) is the same as 2x + 2.
– 2 – 2 2x = x = 4

−3x + 1 = 10 What is different with this problem? We have a negative number of x’s. We will use the white bags to represent the negative x’s. We will also use the idea that –x represents the opposite of x.

Take 1 blue chip from each side.
= −3x + 1 10 Take 1 blue chip from each side.

Divide what is left into 3 equal groups.
= −3x + 1 10 Divide what is left into 3 equal groups.

= −3x + 1 10

= -3x + 1 10 -x = 3 the opposite of x is 3 the opposite of what is 3

= -3x + 1 10 -x = 3 x = -3

-3x + 1 = 10 We took away 1 from each side. -3x + 1 = 10 – 1 – 1 -3 x = 9 Then we divided what was left into 3 equal groups. -3 x = 9

-3x + 1 = 10 We took away 1 from each side. -3x + 1 = 10 – 1 – 1 -3 x = 9 Then we divided what was left into 3 equal groups. -3 x = -x = 3

-3x + 1 = 10 -x = 3 This means the opposite of x is 3; therefore, x must be the opposite of 3, which is -3. x = -3 Once students understand this idea, you can divide both sides by -3 (as a shortcut).

2(x + 1) + 3x – 7 = 5 The cool thing is that we can use these manipulatives for even more complex problems. Using manipulatives is helpful in visualizing what is going on in the problem. The goal is to solve the problem correctly no matter what method is used. Ideally, students will move from using the manipulatives to solving the equations using symbols & procedures.

Do you see all of the parts of the equation?
2(x + 1) + 3x – 7 = 5 Make sure they can see all the parts on the LHS. Do you see all of the parts of the equation?

Pair neutrals or zero pairs
2(x + 1) + 3x – 7 = 5 Pair neutrals or zero pairs

Clear neutrals or zero pairs
2(x + 1) + 3x – 7 = 5 Clear neutrals or zero pairs

2(x + 1) + 3x – 7 = 5 We now have 5x – 5 = 5 Add 5 to both sides

Pair zeros or neutral pairs
2(x + 1) + 3x – 7 = 5 Pair zeros or neutral pairs

Clear zeros or neutral pairs
2(x + 1) + 3x – 7 = 5 Clear zeros or neutral pairs

2(x + 1) + 3x – 7 = 5 5x = 10

2(x + 1) + 3x – 7 = 5 x = 2

2(x + 1) + 3x – 7 = 5 Explain each step. 2(x + 1) + 3x – 7 = 5 2x x – 7 = 5 5x – 5 = x = 10

2(x + 1) + 3x – 7 = 5 Explain each step. 2(x + 1) + 3x – 7 = 5 2x x – 7 = 5 5x – 5 = x = x = 2

Now we have x’s on both sides of the equation. No worries…
4x – 3 = 5x + 2 Now we have x’s on both sides of the equation. No worries…

= 4x – 3 5x + 2 We can take away 4 x’s from both sides.

= 4x – 3 5x + 2 We want to get the x by itself.
The x has 2 blue chips with it.

= 4x – 3 5x + 2 We can’t take away 2 blue chips from both sides because we don’t have 2 blue chips on the LHS. Remember that taking away is the same as adding the opposite. We can add 2 negatives / white chips to both sides.

We can clear neutrals or zero pairs on the RHS.
= 4x – 3 5x + 2 We can clear neutrals or zero pairs on the RHS.

Symmetric property of equality
= 4x – 3 5x + 2 -5 = x x = -5 Symmetric property of equality

4x – 3 = 5x + 2 Explain each step. 4x – 3 = 5x + 2 – 4x – 4x – 3 = x + 2 – 2 – 2 –5 = x x = –5

Back to Solving Equations
standards/8/EE/C/7/tasks/392 Now we are going back to the original problem in this ppt from Illustrative Mathematics. We are going to abstract what we have done with the concrete manipulatives by drawing pictures instead.

2x + 3 = 7

2x + 3 = 7 −3 −3 2x = 4

2x = 4 x = 2

2x = 4 x = 2 From this picture, we can see that, in order to keep the balance level, each bag must contain 2 tiles.

Solve in two ways: 5x + 1 = 2x + 7 symbolically, the way you usually do with equations, and also with pictures of a balance. Show how each step you take symbolically is shown in the pictures.

Solve the equation 4x = x + 1 using pictures and symbols. Discuss any issues that arise.

4x = x + 1 When you solve this equation with pictures, you end up with 3 bags balancing with 1 tile. In order to do the division, you have to cut the tile, leading to the fraction 1/3, which is the solution you get symbolically.

What issues arise when you try to solve the equation 2 = 2x − 4 using pictures? Do the same issues arise when you solve this equation symbolically?

2 = 2x − 4 In order to solve this equation with pictures, you have to have some way of representing the subtraction in 2x – 4. If students have experience with integer chips, they can transfer that knowledge to this situation to show 2x + (−4), but otherwise they may struggle with the idea. The pictures give us a nice model for understanding the operations we do to solve equations, but it is only smooth for problems with “nice” numbers. This is one reason why we want to move to the symbolic approach.

2x + 4 = 10 Use pictures to show why the following solution to the equation 2x + 4 = 10 is incorrect:

2x + 4 = 10 The mistake is in the first step - the student divided only part of the left-hand-side of the equation by 2. You can see in the picture that splitting the equation this way will not keep the balance level (assuming the two bags are equal):

None or Infinitely Many
Make up a linear equation that has no solutions. What would happen if you solved this equation with pictures? How is this different than an equation that has infinitely many solutions?

None or Infinitely Many
A linear equation will have no solution if there are the same number of x’s and different constants on each side. For example: 2x + 4 = 2x + 1 If you solve this with pictures, when you take away the 2x from both sides you will end up with 4 = 1, which clearly cannot be balanced. If the equation had infinitely many solutions, you would find that you had exactly the same picture on the two sides of the balance.

Cover-up the part of the equation with the variable.
Cover-up Method x + 2 = 7 Cover-up the part of the equation with the variable.

Cover-up Method x + 2 = 7 Cover-up the part of the equation with the variable. + 2 = 7 This means “something plus 2 is 7.” So, what plus 2 is 7? 5

Symbolically x + 2 = 7 – 2 – 2 x = 5

Cover-up Method x – 3 = 5

This means “something minus 3 is 5.”
Cover-up Method x – 3 = 5 – 3 = 5 This means “something minus 3 is 5.” So, what minus 3 is 5? 8

Symbolically x – 3 = x = 8

Cover-up Method 2x = 8

This means “2 times something is 8.”
Cover-up Method 2x = 8 = 8 This means “2 times something is 8.” So, 2 times what is 8? 4

Cover-up Method 3x = −6

This means “3 times something is −6.”
Cover-up Method 3x = −6 = −6 This means “3 times something is −6.” So, 3 times what is −6? −2

3x + 2 = 5 Cover-up the part of the equation with the variable.
Cover-up Method 3x + 2 = 5 Cover-up the part of the equation with the variable.

Cover-up Method 3x + 2 = 5 + 2 = 5 What plus 2 is 5? 3
The term covered up by the hand represents a 3. The hand was covering 3x, which must equal 3. 3x = 3

Cover-up Method Now we have 3x = 3 = 3

Cover-up Method Now we have 3x = 3 = 3 This means “3 times something is 3.”

Cover-up Method Now we have 3x = 3 = 3 This means “3 times something is 3.” =1 x = 1

Symbolically 3x + 2 = 5 – 2 – 2 3x = 3

Symbolically 3x + 2 = 5 – 2 – 2 3x = x = 1

Cover-up the part of the equation with the variable.
Cover-up Method 2(x + 1) = 10 Cover-up the part of the equation with the variable.

Cover-up Method 2(x + 1) = 10 2 = 10
Cover-up the part of the equation with the variable. = 10 This means “two times something is 10.” So, 2 times what is 10? 5

The hand was covering (x + 1) which must equal 5. x + 1 = 5
Cover-up Method The hand was covering (x + 1) which must equal 5. x + 1 = 5

The hand was covering (x + 1) which must equal 5. x + 1 = 5 + 1 = 5
Cover-up Method The hand was covering (x + 1) which must equal 5. x + 1 = = 5

Cover-up Method The hand was covering (x + 1) which must equal 5. x + 1 = = 5 = 4 x = 4

Explain each step. 2(x + 1) = 10
Symbolically Explain each step. 2(x + 1) = 10

Symbolically Explain each step. 2(x + 1) = 10 2 2 x + 1 = 5 – 1 – 1
– – 1 x = 4 NOTE: This is a different symbolic strategy than we used before when we were using manipulatives. Before we did the distributive property first.

Back to manipulatives What if we had an equation like y = 2x + 3 We can introduce another set of “bags.” Pink is positive. White w/ pink triangles is negative. The “bag of pebbles” represent an unknown amount – we do not know how many “counters” are inside the bag. Each bag of pebbles represents y or −y.

= y 2x + 3 What should we put here? What should we put here?

= y 2x + 3

Back to manipulatives What if we had an equation like y = 2x + 3? With this kind of problem our goal is usually to solve for y, which means we want to get y on one side of the equation all by itself. There was nothing to do to this problem except to visualize it with counters and bags.

Solve this equation for y. Model with manipulatives.
y – 3 = 4x + 5 Solve this equation for y. Model with manipulatives.

Add 3 positives to both sides.
y – 3 = 4x + 5 Add 3 positives to both sides.

Pair neutrals or zero pairs.
y – 3 = 4x + 5 Pair neutrals or zero pairs.

Clear neutrals or zero pairs.
y – 3 = 4x + 5 Clear neutrals or zero pairs.

y – 3 = 4x + 5 y = 4x + 8

Symbolically y – 3 = 4x y = 4x + 8

Solve this equation for y. Model with manipulatives.
6x + 3y = 9 Solve this equation for y. Model with manipulatives.

6x + 3y = 9 We want to take away the 6 x’s from both sides.
But we don’t have 6 x’s on the RHS. We can add 6 –x’s to both sides.

6x + 3y = 9 Pair neutrals or zero pairs.

6x + 3y = 9 Clear neutrals or zero pairs.

We need to split the RHS into 3 equal groups.
6x + 3y = 9 We now have 3y = –6x + 9. We need to split the RHS into 3 equal groups.

6x + 3y = 9

6x + 3y = 9 y = –2x + 3

Symbolically Explain each step. 6x + 3y = 9 – 6x – 6x 3y = 9 – 6x 3y = – 6x + 9 3y = 3 (– 2x + 3)

Symbolically Explain each step. 6x + 3y = 9 – 6x – 6x 3y = 9 – 6x 3y = – 6x + 9 3y = 3 (–2x + 3) y = –2x + 3

Systems of Equations We can continue these ideas with systems of equations… Solve y = 2x + 3 y = 4x – 7 Simply substitute one y for the other. 2x + 3 = 4x – 7 Don’t forget to solve for y also & write your solution as (x, y).

Algebra Tiles Show these with algebra tiles: x + 2 = 7 3x = −6 3x + 2 = 5 -3x + 1 = 10 2(x + 1) + 3x – 7 = 5 4x – 3 = 5x + 2

Algebra Tiles Multiply these: Use the area model for multiplication. 2 x 3 2(x + 2) (x + 1)(x + 2)

Algebra Tiles Factor these: Turn it in to a multiplication sentence. Make a rectangle with all of the pieces. Find the dimensions of the rectangle. 12 2x + 6 x2 + 5x + 6

Solving Equations Conceputally

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