Math / Algebra

QuestionExercises Write the equation modeled by the algebra tiles. 1.
2. $\square$ Use algebra tiles to model and solve each equation.
3. $x-3=2$
4. $x-4=7$
5. $x+1=5$
6. $x+4=7$
7. $1+x=-3$
8. $5+x=-3$
9. $x-4=-3$
10. $-4+x=-8$

Studdy Solution

STEP 1

What is this asking? We need to write equations based on algebra tiles and then use algebra tiles (not really, just imagine them!) to solve some given equations. Watch out! Don't mix up the positive and negative tiles!
Remember, green represents the variable $x$ , small yellow squares are positive numbers, and small red squares are negative numbers.

STEP 2

1. Write equations from tiles
2. Solve equations with imaginary tiles

STEP 3

Alright, so we've got one long green tile, which means $x$ , and three little yellow squares, meaning $+3$ .
On the other side, we've got five little yellow squares, meaning $+5$ .
So, our equation is $x + 3 = 5$ .
Boom!

STEP 4

Now, we see two long green tiles!
That's $2x$ .
And on the other side, four little yellow squares, which is $+4$ .
So, the equation is $2x = 4$ .
Nailed it!

STEP 5

Imagine $x$ and three red squares (for $-3$ ) on one side, and two yellow squares (for $+2$ ) on the other.
To get $x$ by itself, we need to add three yellow squares to both sides.
This is like adding $+3$ to both sides of the equation: $x - 3 + 3 = 2 + 3$ .
The $-3$ and $+3$ on the left add to zero, leaving us with $x = 5$ .
Awesome!

STEP 6

$x$ and four red squares on one side, seven yellow squares on the other.
Add four yellow squares to both sides: $x - 4 + 4 = 7 + 4$ .
So, $x = 11$ !

STEP 7

$x$ and one yellow square on one side, five yellow squares on the other.
Add one red square to both sides: $x + 1 - 1 = 5 - 1$ .
Therefore, $x = 4$ .

STEP 8

$x$ and four yellow squares on one side, seven yellow squares on the other.
Add four red squares to both sides: $x + 4 - 4 = 7 - 4$ .
Thus, $x = 3$ .

STEP 9

One yellow square and $x$ on one side, three red squares on the other.
Add one red square to both sides: $1 - 1 + x = -3 - 1$ .
So, $x = -4$ .

STEP 10

Five yellow squares and $x$ on one side, three red squares on the other.
Add five red squares to both sides: $5 - 5 + x = -3 - 5$ .
Hence, $x = -8$ .

STEP 11

$x$ and four red squares on one side, three red squares on the other.
Add four yellow squares to both sides: $x - 4 + 4 = -3 + 4$ .
Therefore, $x = 1$ .

STEP 12

Four red squares and $x$ on one side, eight red squares on the other.
Add four yellow squares to both sides: $-4 + 4 + x = -8 + 4$ .
Thus, $x = -4$ .

STEP 13

We wrote equations representing the algebra tiles: $x + 3 = 5$ and $2x = 4$ .
We also solved a bunch of equations using our imaginary algebra tiles, finding the value of $x$ in each case!

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QuestionExercises Write the equation modeled by the algebra tiles. 1.
2. $\square$ Use algebra tiles to model and solve each equation.
3. $x-3=2$
4. $x-4=7$
5. $x+1=5$
6. $x+4=7$
7. $1+x=-3$
8. $5+x=-3$
9. $x-4=-3$
10. $-4+x=-8$

Studdy Solution

STEP 1

STEP 2

1. Write equations from tiles
2. Solve equations with imaginary tiles

STEP 3

Alright, so we've got one long green tile, which means $x$ , and three little yellow squares, meaning $+3$ .
On the other side, we've got five little yellow squares, meaning $+5$ .
So, our equation is $x + 3 = 5$ .
Boom!

STEP 4

Now, we see two long green tiles!
That's $2x$ .
And on the other side, four little yellow squares, which is $+4$ .
So, the equation is $2x = 4$ .
Nailed it!

STEP 5

STEP 6

$x$ and four red squares on one side, seven yellow squares on the other.
Add four yellow squares to both sides: $x - 4 + 4 = 7 + 4$ .
So, $x = 11$ !

STEP 7

$x$ and one yellow square on one side, five yellow squares on the other.
Add one red square to both sides: $x + 1 - 1 = 5 - 1$ .
Therefore, $x = 4$ .

STEP 8

$x$ and four yellow squares on one side, seven yellow squares on the other.
Add four red squares to both sides: $x + 4 - 4 = 7 - 4$ .
Thus, $x = 3$ .

STEP 9

One yellow square and $x$ on one side, three red squares on the other.
Add one red square to both sides: $1 - 1 + x = -3 - 1$ .
So, $x = -4$ .

STEP 10

Five yellow squares and $x$ on one side, three red squares on the other.
Add five red squares to both sides: $5 - 5 + x = -3 - 5$ .
Hence, $x = -8$ .

STEP 11

$x$ and four red squares on one side, three red squares on the other.
Add four yellow squares to both sides: $x - 4 + 4 = -3 + 4$ .
Therefore, $x = 1$ .

STEP 12

Four red squares and $x$ on one side, eight red squares on the other.
Add four yellow squares to both sides: $-4 + 4 + x = -8 + 4$ .
Thus, $x = -4$ .

STEP 13

We wrote equations representing the algebra tiles: $x + 3 = 5$ and $2x = 4$ .
We also solved a bunch of equations using our imaginary algebra tiles, finding the value of $x$ in each case!

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Studdy Solution

STEP 1

STEP 2

1. Write equations from tiles2. Solve equations with imaginary tiles

STEP 3

Alright, so we've got one long green tile, which means xxx, and three little yellow squares, meaning +3+3+3.On the other side, we've got five little yellow squares, meaning +5+5+5.So, our equation is x+3=5x + 3 = 5x+3=5.Boom!

STEP 4

Now, we see *two* long green tiles!That's 2x2x2x.And on the other side, four little yellow squares, which is +4+4+4.So, the equation is 2x=42x = 42x=4.Nailed it!

STEP 5

STEP 6

xxx and four red squares on one side, seven yellow squares on the other.Add four yellow squares to both sides: x−4+4=7+4x - 4 + 4 = 7 + 4x−4+4=7+4.So, x=11x = 11x=11!

STEP 7

xxx and one yellow square on one side, five yellow squares on the other.Add one red square to both sides: x+1−1=5−1x + 1 - 1 = 5 - 1x+1−1=5−1.Therefore, x=4x = 4x=4.

STEP 8

xxx and four yellow squares on one side, seven yellow squares on the other.Add four red squares to both sides: x+4−4=7−4x + 4 - 4 = 7 - 4x+4−4=7−4.Thus, x=3x = 3x=3.

STEP 9

One yellow square and xxx on one side, three red squares on the other.Add one red square to both sides: 1−1+x=−3−11 - 1 + x = -3 - 11−1+x=−3−1.So, x=−4x = -4x=−4.

STEP 10

Five yellow squares and xxx on one side, three red squares on the other.Add five red squares to both sides: 5−5+x=−3−55 - 5 + x = -3 - 55−5+x=−3−5.Hence, x=−8x = -8x=−8.

STEP 11

xxx and four red squares on one side, three red squares on the other.Add four yellow squares to both sides: x−4+4=−3+4x - 4 + 4 = -3 + 4x−4+4=−3+4.Therefore, x=1x = 1x=1.

STEP 12

Four red squares and xxx on one side, eight red squares on the other.Add four yellow squares to both sides: −4+4+x=−8+4-4 + 4 + x = -8 + 4−4+4+x=−8+4.Thus, x=−4x = -4x=−4.

STEP 13

We wrote equations representing the algebra tiles: x+3=5x + 3 = 5x+3=5 and 2x=42x = 42x=4.We also solved a bunch of equations using our imaginary algebra tiles, finding the value of xxx in each case!

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Write equations from tiles
2. Solve equations with imaginary tiles

Alright, so we've got one long green tile, which means $x$ , and three little yellow squares, meaning $+3$ .
On the other side, we've got five little yellow squares, meaning $+5$ .
So, our equation is $x + 3 = 5$ .
Boom!

Now, we see two long green tiles!
That's $2x$ .
And on the other side, four little yellow squares, which is $+4$ .
So, the equation is $2x = 4$ .
Nailed it!

$x$ and four red squares on one side, seven yellow squares on the other.
Add four yellow squares to both sides: $x - 4 + 4 = 7 + 4$ .
So, $x = 11$ !

$x$ and one yellow square on one side, five yellow squares on the other.
Add one red square to both sides: $x + 1 - 1 = 5 - 1$ .
Therefore, $x = 4$ .

$x$ and four yellow squares on one side, seven yellow squares on the other.
Add four red squares to both sides: $x + 4 - 4 = 7 - 4$ .
Thus, $x = 3$ .

One yellow square and $x$ on one side, three red squares on the other.
Add one red square to both sides: $1 - 1 + x = -3 - 1$ .
So, $x = -4$ .

Five yellow squares and $x$ on one side, three red squares on the other.
Add five red squares to both sides: $5 - 5 + x = -3 - 5$ .
Hence, $x = -8$ .

$x$ and four red squares on one side, three red squares on the other.
Add four yellow squares to both sides: $x - 4 + 4 = -3 + 4$ .
Therefore, $x = 1$ .

Four red squares and $x$ on one side, eight red squares on the other.
Add four yellow squares to both sides: $-4 + 4 + x = -8 + 4$ .
Thus, $x = -4$ .

We wrote equations representing the algebra tiles: $x + 3 = 5$ and $2x = 4$ .
We also solved a bunch of equations using our imaginary algebra tiles, finding the value of $x$ in each case!