Question

use the distributive property to write equivalent expressions. use the distributive property to write an equivalent expression. $15x - 12(6 - 7x)$ $= 15x + -12(6 + -7x)$ $= 15x + (\square \cdot \square) + (\square \cdot \square)$

Explanation:

Step1: Recall distributive property

The distributive property is \(a(b + c)=ab + ac\). Here, we have \(-12(6 + (-7x))\), so we apply the distributive property to \(-12\) and the terms inside the parentheses.

Step2: Apply distributive property

For \(-12(6 + (-7x))\), we multiply \(-12\) with \(6\) and \(-12\) with \(-7x\) separately. So it becomes \((-12\cdot6)+(-12\cdot(-7x))\), which is the same as \((-12\cdot6)+(12\cdot7x)\) (since multiplying two negatives gives a positive). But in the form of the equation given, we have \(15x+(-12\cdot6)+(-12\cdot(-7x))\), which can be written as \(15x+(-12\cdot6)+(12\cdot7x)\), but following the structure of the boxes, we use the original multiplication of \(-12\) with \(6\) and \(-12\) with \(-7x\). Wait, actually, the expression is \(15x + -12(6 + -7x)\), so distributing \(-12\) over \(6\) and \(-7x\) gives \(15x+(-12\cdot6)+(-12\cdot(-7x))\). But in the boxed form, we need to fill in the blanks for \((\square\cdot\square)+(\square\cdot\square)\) from the distribution of \(-12\) over \(6\) and \(-7x\). So the first product is \(-12\cdot6\) and the second product is \(-12\cdot(-7x)\), but since the problem has the form \(15x+(\square\cdot\square)+(\square\cdot\square)\), we take the coefficients as they are. So the first two blanks are \(-12\) and \(6\), the next two are \(-12\) and \(-7x\)? Wait, no, let's re - examine. The original expression after rewriting subtraction as addition of negative is \(15x+(-12)(6 + (-7x))\). Then by distributive property \(a(b + c)=ab+ac\), here \(a=-12\), \(b = 6\), \(c=-7x\). So \(ab=-12\times6\) and \(ac=-12\times(-7x)\). So the expression becomes \(15x+(-12\times6)+(-12\times(-7x))\). So in the form \(15x+(\square\cdot\square)+(\square\cdot\square)\), the first product is \(-12\cdot6\) and the second product is \(-12\cdot(-7x)\). But if we consider the signs, \(-12\times(-7x)=84x\), but the problem is just about the distributive step, not simplifying the signs yet. So the blanks should be filled with \(-12\), \(6\), \(-12\), \(-7x\)? Wait, no, the problem has the expression \(15x+(\square\cdot\square)+(\square\cdot\square)\), so from the distribution of \(-12\) over \(6\) and \(-7x\), we have:

\(15x+(-12\cdot6)+(-12\cdot(-7x))\)

So the first pair of blanks is \(-12\) and \(6\), the second pair is \(-12\) and \(-7x\)? But maybe the problem expects us to write the multiplication of \(-12\) with \(6\) and \(-12\) with \(-7x\) as per the distributive property. So the first two boxes are \(-12\) and \(6\), the next two are \(-12\) and \(-7x\). But let's check the arithmetic again.

Wait, the original expression is \(15x-12(6 - 7x)\). We know that \(a-(b - c)=a+(-b + c)\), so \(15x-12(6 - 7x)=15x+(-12)(6+(-7x))\). Now, using the distributive property \(m(n + p)=mn+mp\), where \(m=-12\), \(n = 6\), \(p=-7x\). So \(mn=-12\times6\) and \(mp=-12\times(-7x)\). So the expression is \(15x+(-12\times6)+(-12\times(-7x))\). So in the form \(15x+(\square\cdot\square)+(\square\cdot\square)\), the first product is \(-12\times6\) (so \(\square=-12\), \(\square = 6\)) and the second product is \(-12\times(-7x)\) (so \(\square=-12\), \(\square=-7x\)). But if we consider the problem's requirement, we can also note that when we distribute \(-12\) over \(6\) and \(-7x\), the two products are \(-12\times6\) and \(-12\times(-7x)\). So the blanks should be filled as follows: the first two blanks are \(-12\) and \(6\), the next two are \(-12\) and \(-7x\). But let's check the signs again. If we do the distribution correctly, for \(-12(6 - 7x)\), we can also think of it as \(-12\times6-12\times(-7x)\) (since \(a(b - c)=ab - ac\)), which is the same as \(-12\times6 + 12\times7x\). But in the given form \(15x+(\square\cdot\square)+(\square\cdot\square)\), we need to use the distributive property in the form of adding the two products from distributing \(-12\) over \(6\) and \(-7x\) (after rewriting \(6 - 7x\) as \(6+(-7x)\)). So the correct filling is: the first product is \(-12\times6\) (so \(\square=-12\), \(\square = 6\)) and the second product is \(-12\times(-7x)\) (so \(\square=-12\), \(\square=-7x\)). But if we simplify the second product, \(-12\times(-7x)=84x\), but the problem is just about the distributive step, not simplifying the terms. So the blanks are filled with \(-12\), \(6\), \(-12\), \(-7x\) in order. Wait, but maybe the problem expects positive coefficients? Wait, no, let's re - do the distributive property step by step.

Given \(15x-12(6 - 7x)\)

Step 1: Rewrite subtraction as addition of the opposite: \(15x+(-12)(6+(-7x))\)

Step 2: Apply the distributive property \(a(b + c)=ab + ac\) where \(a=-12\), \(b = 6\), \(c=-7x\)

So \(ab=-12\times6\) and \(ac=-12\times(-7x)\)

So the expression becomes \(15x+(-12\times6)+(-12\times(-7x))\)

So in the form \(15x+(\square\cdot\square)+(\square\cdot\square)\), the first pair of blanks is \(-12\) and \(6\), the second pair is \(-12\) and \(-7x\)

Answer:

The first two blanks are \(-12\) and \(6\), the next two are \(-12\) and \(-7x\). So filling in the boxes: \(15x+(\boldsymbol{-12}\cdot\boldsymbol{6})+(\boldsymbol{-12}\cdot\boldsymbol{-7x})\)