Algebra 1 · Unit 1 · Topic 1.3

Distribution and Equivalent Expressions

Use the distributive property and like terms to rewrite expressions without changing their values, then compare equivalent forms on a graph.

Why rewrite a calculation?

Suppose three identical kits each contain \(2x\) blue pieces and \(5\) red pieces. Counting kit by kit gives \(3(2x+5)\) pieces. Counting all blue pieces and all red pieces separately gives \(6x+15\). The count has not changed; only the grouping has. This is the distributive property:

\[ a(b+c)=ab+ac. \]

The letters stand for real numbers. Multiplication by \(a\) acts on both terms in the sum, not only the first. Since subtraction means adding an opposite, the same reasoning gives

\[ a(b-c)=ab-ac. \]

Two expressions are equivalent when they give the same value for every input allowed in both expressions. Distribution is an algebraic reason that \(3(2x+5)\) and \(6x+15\) are equivalent for every real \(x\); the kit story merely makes the counting visible when \(x\) is a nonnegative whole number. The identity itself is not restricted to kit counts.

A negative factor reaches every term

The original example is a sharper test of the rule. Let \(x\) be any real number and simplify \(-3(2x-5)+4x\). Treat \(-3\) as one signed multiplier. Inside the parentheses are the two terms \(2x\) and \(-5\), so each gets multiplied by \(-3\):

\[ \begin{aligned} -3(2x-5)+4x &=(-3)(2x)+(-3)(-5)+4x \\ &=-6x+15+4x \\ &=(-6+4)x+15 \\ &=-2x+15. \end{aligned} \]

The second product is positive because a negative times a negative is positive. In the third line, \(-6x\) and \(4x\) are like terms: both are multiples of the same variable to the same power. Their coefficients add to \(-2\). The constant \(15\) does not combine with an \(x\)-term. We have proved the identity

\[ -3(2x-5)+4x=-2x+15\qquad\text{for every real }x. \]

One tempting shortcut is to distribute the \(-3\) only to \(2x\), producing \(-6x-5+4x=-2x-5\). That misses the product with \(-5\); at \(x=0\), the original expression is \((-3)(-5)=15\), whereas the shortcut gives \(-5\). A single failed input disproves equivalence.

The graph checks the identity, but does not prove it

The table evaluates the original expression \(F(x)=-3(2x-5)+4x\) and the simplified expression \(G(x)=-2x+15\) at four inputs. Here \(x\) and the outputs are abstract number values, not measured quantities.

\(x\)	\(F(x)=-3(2x-5)+4x\)	\(G(x)=-2x+15\)
\(0\)	\(15\)	\(15\)
\(2\)	\(-3(-1)+8=11\)	\(11\)
\(5\)	\(-3(5)+20=5\)	\(5\)
\(8\)	\(-3(11)+32=-1\)	\(-1\)

Original coordinate graph with horizontal x input from zero to eight and vertical expression value from negative two to sixteen. Gold dashed and teal solid plots for negative three times the quantity two x minus five plus four x and negative two x plus fifteen lie exactly on the same descending line. Four table points are marked: zero comma fifteen, two comma eleven, five comma five, and eight comma negative one. — Both forms lie on the same line for the shown interval. The gold dashed plot of \(F\) and teal solid plot of \(G\) are drawn on top of each other; dots mark the table values.

At \(x=2\), for example, the graph's height is \(11\) for either form. Seeing matching outputs at four inputs is a useful error check, but four checks alone cannot establish that two expressions agree at every input. The distributive calculation above establishes that broader claim. Conversely, if even one pair of graph points had different heights at the same \(x\), the expressions would not be equivalent.

The line continues beyond the displayed interval because \(-2x+15\) is defined for every real \(x\). The graph shows only \(0\le x\le8\), so the picture is a window into the identity, not a restriction on where it holds.

Subtraction can hide a factor of negative one

A minus sign before parentheses is easy to overlook because the multiplier \(-1\) is invisible. Write it explicitly when signs are uncertain:

\[ 8-(3x-2)=8+(-1)(3x-2)=8-3x+2=10-3x. \]

The constant becomes \(+2\), not \(-2\). As a check, at \(x=0\) the original expression is \(8-(0-2)=10\), which agrees with \(10-3(0)=10\).

Distribution also works backward. In \(6x+9\), both terms have a factor of \(3\), so \(6x+9=3(2x+3)\). Expanding the right side returns \(6x+9\). This reverse check is especially useful when an expression has several terms: if re-expansion does not recover every signed term, the proposed rewrite is not equivalent.