Mathematical Process Standards>Justifying Mathematical Ideas with Precise Language(TEKS.Math.8.1.G)
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Texas 8th Grade Math › Mathematical Process Standards>Justifying Mathematical Ideas with Precise Language(TEKS.Math.8.1.G)
Claim: For any two distinct points on the graph of $y=3x+1$, the slope between them is $3$. Which explanation best justifies this claim?
Choose two points on the line, such as $(0,1)$ and $(1,4)$. By the slope formula, $m=\frac{4-1}{1-0}=3$, so the slope is $3$.
Let two generic points on the line be $(x_1,,3x_1+1)$ and $(x_2,,3x_2+1)$ with $x_2\ne x_1$. By the slope formula, $$m=\frac{(3x_2+1)-(3x_1+1)}{x_2-x_1}=\frac{3(x_2-x_1)}{x_2-x_1}=3,$$ using factoring and cancellation (multiplicative inverses).
In $y=3x+1$, the $1$ is the $x$-coefficient, so it must be the slope.
Divide both sides of $y=3x+1$ by $x$ to get $\frac{y}{x}=3+\frac{1}{x}$; the $3$ is the slope since it is next to $x$ after dividing.
Explanation
B uses the slope formula with arbitrary points $(x_1,3x_1+1)$ and $(x_2,3x_2+1)$, then factors and cancels $(x_2-x_1)$ to conclude $m=3$, correctly naming the slope formula, factoring, and multiplicative inverses. A verifies only one example and does not justify the claim for all points. C misidentifies the slope: in $y=mx+b$, the slope is the coefficient of $x$ (here $3$), not the constant term $1$. D performs an invalid step for determining slope; dividing by $x$ produces $\frac{y}{x}$, which is not the definition of slope between two points and varies with $x$.
Claim: $4(x-3)+2x$ is equivalent to $6x-12$. Which explanation best justifies this claim?
Factor out $2$: $4(x-3)+2x=2\big(2(x-3)+x\big)=2(2x-6+x)=2(3x-6)=6x-6$; this matches.
Subtract inside first: $x-3=1x$, so $4(1x)+2x=6x$, and the $-12$ cancels because $+2x$ balances it.
Distribute $4$ to $x$ only: $4(x-3)+2x=4x-3+2x=6x-3$; then lower the constant to $-12$ because there was a $-3$ three times.
Apply the distributive property and combine like terms: $4(x-3)+2x=4x-12+2x=(4x+2x)-12=6x-12$, using distribution and combining like terms.
Explanation
D correctly uses the distributive property to get $4x-12$ and then combines like terms $4x$ and $2x$ to obtain $6x-12$. A's factorization changes the expression incorrectly and yields $6x-6$, which is not equivalent. B incorrectly treats $x-3$ as $1x$ and ignores the constant term $-12$. C misapplies distribution (omits multiplying $-3$ by $4$) and then makes an unjustified adjustment to the constant.
Claim: The equation $3(2y-5)=6y-15$ is true for all real $y$. Which explanation best justifies this claim?
Use the distributive property on the left: $3(2y-5)=6y-15$, so the equation becomes $6y-15=6y-15$. Subtracting the right side from both sides gives $0=0$, an identity; therefore it holds for all $y$.
Divide both sides by $y$ to get $6-\frac{15}{y}=6-\frac{15}{y}$. Since both sides match, the equation is always true.
Test $y=5$: $3(10-5)=15$ and $6(5)-15=15$. It works once, so it is true for all $y$.
Because $3$ and $2y-5$ are both factors on the left, the product equals the right side only when $y$ is even, which means it works for all real $y$.
Explanation
A correctly expands the left side to $6y-15$, then notes $6y-15=6y-15$ and simplifies to $0=0$, an identity, so the equation is true for all real $y$. B divides by $y$, which is invalid when $y=0$ and unnecessary; it does not justify truth for all $y$. C provides only a single example, which is insufficient to prove the statement for all $y$. D makes an unsupported claim about parity and does not use valid algebraic reasoning.
Claim: The system $\begin{cases} y=2x+1 \\ 4x-2y=-2 \end{cases}$ has infinitely many solutions. Which explanation best justifies this claim?
Solve the first equation for $x$: $x=\frac{y-1}{2}$, and substitute into $4x-2y=-2$ to find a single ordered pair that works.
Compare slopes: the first line has slope $2$; the second has slope $-2$. Different slopes imply infinitely many solutions.
Multiply $y=2x+1$ by $-2$ to get $-2y=-4x-2$, which rearranges to $4x-2y=-2$. The second equation is a scalar multiple of the first, so both equations describe the same line and there are infinitely many solutions.
Graph both: they meet at the $y$-intercept $1$, so there is exactly one solution.
Explanation
C shows the second equation is obtained by multiplying the first by $-2$, so the equations are equivalent and represent the same line; thus, every point on the line satisfies both, yielding infinitely many solutions. A incorrectly claims a single solution after substitution. B miscomputes the second line's slope; rewriting $4x-2y=-2$ as $y=2x+1$ shows the same slope and intercept. D asserts a single intersection without justification; identical lines coincide at all points, not just one.
Claim: $(x+5)(x-5)=x^2-25$. Which explanation best justifies this claim?
Square each part and add: $x^2+5^2=x^2+25$, which matches.
Use distribution (FOIL) and additive inverses: $(x+5)(x-5)=x^2-5x+5x-25=x^2-25$; the middle terms cancel because $-5x$ and $+5x$ are additive inverses.
Because $x+5$ and $x-5$ are additive inverses, their product must be $0$, so only $x^2-25$ can match.
Check one value, $x=5$: $(5+5)(5-5)=10\cdot 0=0$ and $5^2-25=0$, so the identity holds.
Explanation
B correctly multiplies binomials via distribution (FOIL), then uses the additive inverse property to cancel $-5x$ and $+5x$, yielding $x^2-25$. A incorrectly applies squaring to a product of sums. C confuses additive inverses with multiplicative inverses; $(x+5)$ and $(x-5)$ are not additive inverses, and their product is not $0$ in general. D verifies only one value, which is insufficient to justify an identity.