The Distributive Property for 7th Grade: Using a rectangle

The area of the rectangle is equal to the length multiplied by the width.

We begin by writing the exercise according to the existing data:

$7\times(4+5)$

We then solve the exercise by using the distributive property, that is, we multiply 7 by each of the terms inside of the parentheses:

$(7\times4)+(7\times5)=$

Lastly we solve the exercise inside of the parentheses and obtain the following:

$28+35=63$

The area of a rectangle is equal to its length multiplied by the width.

We begin by writing the following exercise using the data shown in the figure:

$(3+2)\times(9+4)=$

We solve the exercise using the distributive property.

That is:

We multiply the first term of the left parenthesis by the first term of the right parenthesis.

We then multiply the first term of the left parenthesis by the second term of the right parenthesis.

Now we multiply the second term of the left parenthesis by the first term of the left parenthesis.

Finally, we multiply the second term of the left parenthesis by the second term of the right parenthesis.

In the following way:

$(3\times9)+(3\times4)+(2\times9)+(2\times4)=$

We solve each of the exercises within the parentheses:

$27+12+18+8=$

Lastly we solve the exercise from left to right:

$27+12=39$

$39+18=57$

$57+8=65$

We know that the area of a rectangle is equal to its length multiplied by its width.

We begin by writing an equation with the available data.

$78=3\times(x+7)$

We then use the distributive property to solve the equation.

That is, we multiply each of the terms inside of the parentheses by 3:

$78=3\times x+3\times7$

$78=3x+21$

We move 21 to the other side and use the appropriate sign:

$78-21=3x$

$57=3x$

Lastly we divide both sides by 3:

$\frac{57}{3}=\frac{3x}{3}$

$x=19$

We know that the area of a rectangle is equal to its length multiplied by its width.

We begin by writing an equation with the available data.

$(4x+x^2)\times(3x+8+5x)$

Next we use the distributive property to solve the equation.

$(4x\times3x)+(4x\times8)+(4x\times5x)+(x^2\times3x)+(x^2\times8)+(x^2\times5x)=$

We then solve each of the exercises within the parentheses:

$12x^2+32x+20x^2+3x^3+16x^2+5x^3=$

Finally we add up all the coefficients of X squared and all the coefficients of X cubed and we obtain the following:

$48x^2+8x^3+32x$

In order to solve the exercise, we first need to know the total area of the fence.

Let's remember that the area of a rectangle equals length times width.

Let's write the exercise according to the given data:

$7x\times(30x+4)$$$

We'll use the distributive property to solve the exercise. That means we'll multiply 7x by each term in the parentheses:

$(7x\times30x)+(7x\times4)=$

Let's solve each term in the parentheses and we'll get:

$210x^2+28x$

Now to calculate the painting time, we'll use the formula:

$\frac{7m^2}{\frac{1}{2}hr}=14\frac{m^2}{hr}$

The time will be equal to the area divided by the work rate, meaning:

$\frac{210x^2+28x}{14}$

Let's separate the exercise into addition between fractions:

$\frac{210x^2}{14}+\frac{28x}{14}=$

We'll reduce by 14 and get:

$15x^2+2x$

And this is Isaac's work time.

Let's remember that the area of a rectangle equals length times width.

Let's write the exercise according to the data shown in the drawing:

$(x+9)\times(3x+5)=$

We'll solve the exercise using the distributive property.

That is:

Multiply the first term in the left parentheses by the first term in the right parentheses,

Multiply the first term in the left parentheses by the second term in the right parentheses,

Multiply the second term in the left parentheses by the first term in the right parentheses,

Multiply the second term in the left parentheses by the second term in the right parentheses.

As follows:

$(x\times3x)+(x\times5)+(9\times3x)+(9\times5)=$

Let's solve what's in the parentheses:

$3x^2+5x+27x+45=$

Let's combine like terms with x and we get:

$3x^2+32x+45$

Let's check if there's another expression from the list that could match the expression we got.

Note that we can write the expression in another way, by factoring out x and 9 like this:

$x(3x+5)+9(3x+5)$

If we multiply x and 9 by each term in the parentheses we get:

$(3x^2+5x)+(27x+45)=$

Which is actually the same equation we got before.

Therefore, the matching expressions are the third expression and the sixth expression.