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Common Core: 8th Grade Math : Expressions & Equations

Study concepts, example questions & explanations for Common Core: 8th Grade Math

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All Common Core: 8th Grade Math Resources

7 Diagnostic Tests 75 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

The equation of a line is $\displaystyle 12x-8y=6$. Find the slope of this line.

Possible Answers:

$-\frac{3}{2}$ $\displaystyle -\frac{3}{2}$

$-\frac{3}{4}$ $\displaystyle -\frac{3}{4}$

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

$\frac{3}{4}$ $\displaystyle \frac{3}{4}$

Correct answer:

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

Explanation:

To find the slope, you will need to put the equation in y=mx+b $\displaystyle y=mx+b$ form. The value of $\displaystyle m$ will be the slope.

$\displaystyle 12x-8y=6$

Subtract $\displaystyle 6$ from either side:

$\displaystyle 8y=12x-6$

Divide each side by $\displaystyle 8$:

$y=\frac{3}{2}x-\frac{3}{4}$ $\displaystyle y=\frac{3}{2}x-\frac{3}{4}$

You can now easily identify the value of $\displaystyle m$.

$m=\frac{3}{2}$ $\displaystyle m=\frac{3}{2}$

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Example Question #1 : How To Find X Or Y Intercept

What is the $\displaystyle y$-intercept of the graph of the function $f \left ( x\right ) = 2x^{2} - 7x + 5$ $\displaystyle f \left ( x\right ) = 2x^{2} - 7x + 5$ ?

Possible Answers:

$\left (0,-7 \right )$ $\displaystyle \left (0,-7 \right )$

$\left (0,2 \right )$ $\displaystyle \left (0,2 \right )$

$\left (0,0 \right )$ $\displaystyle \left (0,0 \right )$

$\left (0,5 \right )$ $\displaystyle \left (0,5 \right )$

$\left ( 0, -3 \frac{1}{2}\right )$ $\displaystyle \left ( 0, -3 \frac{1}{2}\right )$

Correct answer:

$\left (0,5 \right )$ $\displaystyle \left (0,5 \right )$

Explanation:

The $\displaystyle y$-intercept of the graph of a function is the point at which it intersects the $\displaystyle y$-axis - that is, at which x = 0 $\displaystyle x = 0$. This point is $\left (0, f(0) \right )$ $\displaystyle \left (0, f(0) \right )$, so evaluate f(0) $\displaystyle f(0)$:

$f \left ( x\right ) = 2x^{2} - 7x + 5$ $\displaystyle f \left ( x\right ) = 2x^{2} - 7x + 5$

$f \left (0\right ) = 2 \cdot 0^{2} - 7 \cdot 0 + 5$ $\displaystyle f \left (0\right ) = 2 \cdot 0^{2} - 7 \cdot 0 + 5$

$f \left (0\right ) = 0- 0 + 5 = 5$ $\displaystyle f \left (0\right ) = 0- 0 + 5 = 5$

The $\displaystyle y$-intercept is $\left (0,5 \right )$ $\displaystyle \left (0,5 \right )$.

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Example Question #2 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

Give the $\displaystyle y$-intercept, if there is one, of the graph of the equation

$y = \frac{1}{x^{2}+ x} + 3$ $\displaystyle y = \frac{1}{x^{2}+ x} + 3$

Possible Answers:

(0,4) $\displaystyle (0,4)$

The graph has no $\displaystyle y$-intercept.

$\left ( 0, \frac{3}{2}\right )$ $\displaystyle \left ( 0, \frac{3}{2}\right )$

(0,2) $\displaystyle (0,2)$

(0,3) $\displaystyle (0,3)$

Correct answer:

The graph has no $\displaystyle y$-intercept.

Explanation:

The $\displaystyle y$-intercept is the point at which the graph crosses the $\displaystyle y$-axis; at this point, the $\displaystyle x$-coordinate is 0, so substitute $\displaystyle 0$ for $\displaystyle x$ in the equation:

$y = \frac{1}{x^{2}+ x} + 3$ $\displaystyle y = \frac{1}{x^{2}+ x} + 3$

$y = \frac{1}{0^{2}+0} + 3$ $\displaystyle y = \frac{1}{0^{2}+0} + 3$

$y = \frac{1}{0} + 3$ $\displaystyle y = \frac{1}{0} + 3$

However, since this expression has 0 in a denominator, it is of undefined value. This means that there is no value of $\displaystyle x$ paired with $\displaystyle y$-coordinate 0, and, subsequently, the graph of the equation has no $\displaystyle y$-intercept.

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Example Question #2 : How To Find X Or Y Intercept

Give the $\displaystyle y$-intercept, if there is one, of the graph of the equation

$y = \frac{3}{\left | 3x-8\right |}$ $\displaystyle y = \frac{3}{\left | 3x-8\right |}$

Possible Answers:

$\left ( 0,- \frac{3}{5}\right )$ $\displaystyle \left ( 0,- \frac{3}{5}\right )$

$\left ( 0, \frac{3}{5}\right )$ $\displaystyle \left ( 0, \frac{3}{5}\right )$

$\left ( 0, -\frac{3}{8}\right )$ $\displaystyle \left ( 0, -\frac{3}{8}\right )$

The graph has no $\displaystyle y$-intercept.

$\left ( 0, \frac{3}{8}\right )$ $\displaystyle \left ( 0, \frac{3}{8}\right )$

Correct answer:

$\left ( 0, \frac{3}{8}\right )$ $\displaystyle \left ( 0, \frac{3}{8}\right )$

Explanation:

$y = \frac{3}{\left | 3x-8\right |}$ $\displaystyle y = \frac{3}{\left | 3x-8\right |}$

$y = \frac{3}{\left | 3 \cdot 0 -8\right |}$ $\displaystyle y = \frac{3}{\left | 3 \cdot 0 -8\right |}$

$y = \frac{3}{\left | 0 -8\right |}$ $\displaystyle y = \frac{3}{\left | 0 -8\right |}$

$y = \frac{3}{\left | -8\right |}$ $\displaystyle y = \frac{3}{\left | -8\right |}$

$y = \frac{3}{ 8}$ $\displaystyle y = \frac{3}{ 8}$

The $\displaystyle y$-intercept is $\left ( 0, \frac{3}{8}\right )$ $\displaystyle \left ( 0, \frac{3}{8}\right )$.

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Example Question #3 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

Give the $\displaystyle y$-intercept, if there is one, of the graph of the equation

$y = 3(x-2)^{2} + 4 (x-7)$ $\displaystyle y = 3(x-2)^{2} + 4 (x-7)$.

Possible Answers:

$\displaystyle (0, -16)$

(0, 8) $\displaystyle (0, 8)$

The graph does not have a $\displaystyle y$-intercept.

$\displaystyle (0, -42)$

$\displaystyle (0, -64)$

Correct answer:

$\displaystyle (0, -16)$

Explanation:

$y = 3(x-2)^{2} + 4 (x-7)$ $\displaystyle y = 3(x-2)^{2} + 4 (x-7)$

$y = 3(0-2)^{2} + 4 (0-7)$ $\displaystyle y = 3(0-2)^{2} + 4 (0-7)$

$y = 3( -2)^{2} + 4 ( -7)$ $\displaystyle y = 3( -2)^{2} + 4 ( -7)$

$y = 3 \cdot 4 + 4 ( -7)$ $\displaystyle y = 3 \cdot 4 + 4 ( -7)$

$\displaystyle y = 12+ 4 ( -7)$

$\displaystyle y = 12+ (-28)$

y = -16 $\displaystyle y = -16$

The $\displaystyle y$-intercept is the point $\displaystyle (0, -16)$.

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Example Question #4 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

A line passes through (5, 2) $\displaystyle (5, 2)$ and is perpendicular to the line of the equation $\displaystyle 3x-4y = 17$. Give the $\displaystyle x$-intercept of this line.

Possible Answers:

The line has no $\displaystyle x$-intercept.

$\left (2 \frac{1}{3}, 0 \right )$ $\displaystyle \left (2 \frac{1}{3}, 0 \right )$

$\left ( -1\frac{3}{4}, 0\right )$ $\displaystyle \left ( -1\frac{3}{4}, 0\right )$

$\left ( 8\frac{2}{3}, 0\right )$ $\displaystyle \left ( 8\frac{2}{3}, 0\right )$

$\left (6 \frac{1}{2}, 0 \right )$ $\displaystyle \left (6 \frac{1}{2}, 0 \right )$

Correct answer:

$\left (6 \frac{1}{2}, 0 \right )$ $\displaystyle \left (6 \frac{1}{2}, 0 \right )$

Explanation:

First, find the slope of the second line $\displaystyle 3x-4y = 17$ by solving for $\displaystyle y$ as follows:

$\displaystyle 3x-4y = 17$

$\displaystyle 3x-4y +4y - 17 = 17+4y - 17$

$\displaystyle 4y = 3x-17$

$\frac{4y }{4}=\frac{ 3x-17 }{4}$ $\displaystyle \frac{4y }{4}=\frac{ 3x-17 }{4}$

$y = \frac{3}{4}x- \frac{17}{4}$ $\displaystyle y = \frac{3}{4}x- \frac{17}{4}$

The equation is now in the slope-intercept form $\displaystyle y = mx+ b$; the slope of the second line is the $\displaystyle x$-coefficient $m= \frac{3}{4}$ $\displaystyle m= \frac{3}{4}$.

The first line, being perpendicular to the second, has as its slope the opposite of the reciprocal of $\frac{3}{4}$ $\displaystyle \frac{3}{4}$, which is $m=- \frac{4}{3}$ $\displaystyle m=- \frac{4}{3}$.

Therefore, we are looking for a line through (5, 2) $\displaystyle (5, 2)$ with slope $m=- \frac{4}{3}$ $\displaystyle m=- \frac{4}{3}$. Using point-slope form

$y - y_{1} = m(x -x_{1})$ $\displaystyle y - y_{1} = m(x -x_{1})$

with

$x _{1 } =5, y _{1 } = 2,m=- \frac{4}{3}$ $\displaystyle x _{1 } =5, y _{1 } = 2,m=- \frac{4}{3}$,

the equation becomes

$y - 2 =- \frac{4}{3}(x -5)$ $\displaystyle y - 2 =- \frac{4}{3}(x -5)$.

To find the $\displaystyle x$-intercept, substitute 0 for $\displaystyle y$ and solve for $\displaystyle x$:

$0 - 2 =- \frac{4}{3}(x -5)$ $\displaystyle 0 - 2 =- \frac{4}{3}(x -5)$

$- 2 =- \frac{4}{3}x+ \frac{20}{3}$ $\displaystyle - 2 =- \frac{4}{3}x+ \frac{20}{3}$

$- 2- \frac{20}{3} =- \frac{4}{3}x+ \frac{20}{3} - \frac{20}{3}$ $\displaystyle - 2- \frac{20}{3} =- \frac{4}{3}x+ \frac{20}{3} - \frac{20}{3}$

$- \frac{26}{3} =- \frac{4}{3}x$ $\displaystyle - \frac{26}{3} =- \frac{4}{3}x$

$- \frac{3} {4}\cdot\left (- \frac{26}{3} \right )=- \frac{3} {4}\cdot \left (- \frac{4}{3}x \right )$ $\displaystyle - \frac{3} {4}\cdot\left (- \frac{26}{3} \right )=- \frac{3} {4}\cdot \left (- \frac{4}{3}x \right )$

$x= \frac{26}{4} = \frac{13}{2} = 6\frac{1}{2}$ $\displaystyle x= \frac{26}{4} = \frac{13}{2} = 6\frac{1}{2}$

The $\displaystyle x$-intercept is the point $\left (6 \frac{1}{2}, 0 \right )$ $\displaystyle \left (6 \frac{1}{2}, 0 \right )$.

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Example Question #273 : Geometry

A line passes through (5, 2) $\displaystyle (5, 2)$ and is parallel to the line of the equation $\displaystyle 3x-4y = 17$. Give the $\displaystyle x$-intercept of this line.

Possible Answers:

$\left ( 8\frac{2}{3}, 0\right )$ $\displaystyle \left ( 8\frac{2}{3}, 0\right )$

The line has no $\displaystyle x$-intercept.

$\left (2 \frac{1}{3}, 0 \right )$ $\displaystyle \left (2 \frac{1}{3}, 0 \right )$

$\left ( -1\frac{3}{4}, 0\right )$ $\displaystyle \left ( -1\frac{3}{4}, 0\right )$

$\left (6 \frac{1}{2}, 0 \right )$ $\displaystyle \left (6 \frac{1}{2}, 0 \right )$

Correct answer:

$\left (2 \frac{1}{3}, 0 \right )$ $\displaystyle \left (2 \frac{1}{3}, 0 \right )$

Explanation:

First, find the slope of the second line $\displaystyle 3x-4y = 17$ by solving for $\displaystyle y$ as follows:

$\displaystyle 3x-4y = 17$

$\displaystyle 3x-4y +4y - 17 = 17+4y - 17$

$\displaystyle 4y = 3x-17$

$\frac{4y }{4}=\frac{ 3x-17 }{4}$ $\displaystyle \frac{4y }{4}=\frac{ 3x-17 }{4}$

$y = \frac{3}{4}x- \frac{17}{4}$ $\displaystyle y = \frac{3}{4}x- \frac{17}{4}$

The equation is now in the slope-intercept form $\displaystyle y = mx+ b$; the slope of the second line is the $\displaystyle x$-coefficient $m= \frac{3}{4}$ $\displaystyle m= \frac{3}{4}$.

The first line, being parallel to the second, has the same slope.

Therefore, we are looking for a line through (5, 2) $\displaystyle (5, 2)$ with slope $m= \frac{3}{4}$ $\displaystyle m= \frac{3}{4}$. Using point-slope form

$y - y_{1} = m(x -x_{1})$ $\displaystyle y - y_{1} = m(x -x_{1})$

with

$x _{1 } =5, y _{1 } = 2,m= \frac{3}{4}$ $\displaystyle x _{1 } =5, y _{1 } = 2,m= \frac{3}{4}$,

the equation becomes

$y - 2 = \frac{3}{4}(x -5)$ $\displaystyle y - 2 = \frac{3}{4}(x -5)$.

To find the $\displaystyle x$-intercept, substitute 0 for $\displaystyle y$ and solve for $\displaystyle x$:

$0 - 2 = \frac{3}{4}(x -5)$ $\displaystyle 0 - 2 = \frac{3}{4}(x -5)$

$- 2 = \frac{3}{4} x - \frac{15}{4}$ $\displaystyle - 2 = \frac{3}{4} x - \frac{15}{4}$

$- 2 + \frac{15}{4}= \frac{3}{4} x - \frac{15}{4} + \frac{15}{4}$ $\displaystyle - 2 + \frac{15}{4}= \frac{3}{4} x - \frac{15}{4} + \frac{15}{4}$

$\frac{3}{4} x = \frac{7}{4}$ $\displaystyle \frac{3}{4} x = \frac{7}{4}$

$\frac{4} {3}\cdot \frac{3}{4} x =\frac{4} {3}\cdot \frac{7}{4}$ $\displaystyle \frac{4} {3}\cdot \frac{3}{4} x =\frac{4} {3}\cdot \frac{7}{4}$

$x = \frac{7}{3} = 2\frac{1}{3}$ $\displaystyle x = \frac{7}{3} = 2\frac{1}{3}$

The $\displaystyle x$-intercept is the point $\left (2 \frac{1}{3}, 0 \right )$ $\displaystyle \left (2 \frac{1}{3}, 0 \right )$.

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Example Question #274 : Geometry

Give the $\displaystyle y$-intercept of the line with slope $\frac{2}{5}$ $\displaystyle \frac{2}{5}$ that passes through point (5, 9) $\displaystyle (5, 9)$.

Possible Answers:

$\left (0, -27 \frac{1}{2} \right )$ $\displaystyle \left (0, -27 \frac{1}{2} \right )$

(0, 7) $\displaystyle (0, 7)$

$\left (0, -17 \frac{1}{2} \right )$ $\displaystyle \left (0, -17 \frac{1}{2} \right )$

The line has no $\displaystyle y$-intercept.

$\displaystyle (0, -11)$

Correct answer:

(0, 7) $\displaystyle (0, 7)$

Explanation:

By the point-slope formula, this line has the equation

$y - y_{1} = m(x -x_{1})$ $\displaystyle y - y_{1} = m(x -x_{1})$

where

$x_{1} = 5, y_{1} = 9 ,m = \frac{2}{5}$ $\displaystyle x_{1} = 5, y_{1} = 9 ,m = \frac{2}{5}$

By substitution, the equation becomes

$y -9 = \frac{2}{5}(x -5)$ $\displaystyle y -9 = \frac{2}{5}(x -5)$

To find the $\displaystyle y$-intercept, substitute 0 for $\displaystyle x$ and solve for $\displaystyle y$:

$y -9 = \frac{2}{5}(0 -5)$ $\displaystyle y -9 = \frac{2}{5}(0 -5)$

$y -9 = \frac{2}{5}( -5)$ $\displaystyle y -9 = \frac{2}{5}( -5)$

$\displaystyle y - 9 = -2$

y = 7 $\displaystyle y = 7$

The $\displaystyle y$-intercept is the point (0, 7) $\displaystyle (0, 7)$.

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Example Question #275 : Geometry

Give the $\displaystyle y$-intercept of the line that passes through points (-3, 4) $\displaystyle (-3, 4)$ and (2, -3) $\displaystyle (2, -3)$.

Possible Answers:

$\left (0, - 5\frac{6}{7} \right )$ $\displaystyle \left (0, - 5\frac{6}{7} \right )$

$\left (0, - \frac{1}{5} \right )$ $\displaystyle \left (0, - \frac{1}{5} \right )$

$\left (0, - \frac{1}{7} \right )$ $\displaystyle \left (0, - \frac{1}{7} \right )$

The line has no $\displaystyle y$-intercept.

$\left (0, - 8 \frac{1}{5} \right )$ $\displaystyle \left (0, - 8 \frac{1}{5} \right )$

Correct answer:

$\left (0, - \frac{1}{5} \right )$ $\displaystyle \left (0, - \frac{1}{5} \right )$

Explanation:

First, find the slope of the line, using the slope formula

$m = \frac{y_{2} -y_{1}}{x_{2}-x_{1}}$ $\displaystyle m = \frac{y_{2} -y_{1}}{x_{2}-x_{1}}$

setting $x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$ $\displaystyle x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$:

$m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$ $\displaystyle m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$

By the point-slope formula, this line has the equation

$y - y_{1} = m(x -x_{1})$ $\displaystyle y - y_{1} = m(x -x_{1})$

where

$x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$ $\displaystyle x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$; the line becomes

$y - 4 = - \frac{7}{5}(x -(-3))$ $\displaystyle y - 4 = - \frac{7}{5}(x -(-3))$

$y - 4 = - \frac{7}{5}(x +3)$ $\displaystyle y - 4 = - \frac{7}{5}(x +3)$

To find the $\displaystyle y$-intercept, substitute 0 for $\displaystyle x$ and solve for $\displaystyle y$:

$y - 4 = - \frac{7}{5}(0+3)$ $\displaystyle y - 4 = - \frac{7}{5}(0+3)$

$y - 4 = - \frac{7}{5}(3)$ $\displaystyle y - 4 = - \frac{7}{5}(3)$

$y - 4 = - \frac{21}{5}$ $\displaystyle y - 4 = - \frac{21}{5}$

$y - 4+ 4 = - \frac{21}{5} + 4$ $\displaystyle y - 4+ 4 = - \frac{21}{5} + 4$

$y = - \frac{1}{5}$ $\displaystyle y = - \frac{1}{5}$

The $\displaystyle y$-intercept is $\left (0, - \frac{1}{5} \right )$ $\displaystyle \left (0, - \frac{1}{5} \right )$.

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Example Question #498 : Ssat Upper Level Quantitative (Math)

Give the $\displaystyle x$-intercept of the line that passes through points (-3, 4) $\displaystyle (-3, 4)$ and (2, -3) $\displaystyle (2, -3)$.

Possible Answers:

$\left ( - 8 \frac{1}{5},0 \right )$ $\displaystyle \left ( - 8 \frac{1}{5},0 \right )$

$\left ( - \frac{1}{5},0 \right )$ $\displaystyle \left ( - \frac{1}{5},0 \right )$

$\left (- 5\frac{6}{7} , 0 \right )$ $\displaystyle \left (- 5\frac{6}{7} , 0 \right )$

$\left (- \frac{1}{7} , 0 \right )$ $\displaystyle \left (- \frac{1}{7} , 0 \right )$

$\text{The line has no } x\text{-intercept.}$ $\displaystyle \text{The line has no } x\text{-intercept.}$

Correct answer:

$\left (- \frac{1}{7} , 0 \right )$ $\displaystyle \left (- \frac{1}{7} , 0 \right )$

Explanation:

First, find the slope of the line, using the slope formula

$m = \frac{y_{2} -y_{1}}{x_{2}-x_{1}}$ $\displaystyle m = \frac{y_{2} -y_{1}}{x_{2}-x_{1}}$

setting $x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$ $\displaystyle x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$:

$m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$ $\displaystyle m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$

By the point-slope formula, this line has the equation

$y - y_{1} = m(x -x_{1})$ $\displaystyle y - y_{1} = m(x -x_{1})$

where

$x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$ $\displaystyle x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$; the line becomes

$y - 4 = - \frac{7}{5}(x -(-3))$ $\displaystyle y - 4 = - \frac{7}{5}(x -(-3))$

$y - 4 = - \frac{7}{5}(x +3)$ $\displaystyle y - 4 = - \frac{7}{5}(x +3)$

To find the $\displaystyle x$-intercept, substitute 0 for $\displaystyle y$ and solve for $\displaystyle x$:

$0 - 4 = - \frac{7}{5}(x +3)$ $\displaystyle 0 - 4 = - \frac{7}{5}(x +3)$

$- 4 = - \frac{7}{5} x - \frac{21}{5}$ $\displaystyle - 4 = - \frac{7}{5} x - \frac{21}{5}$

$- 4+ \frac{21}{5} = - \frac{7}{5} x - \frac{21}{5}+ \frac{21}{5}$ $\displaystyle - 4+ \frac{21}{5} = - \frac{7}{5} x - \frac{21}{5}+ \frac{21}{5}$

$\frac{1}{5} = - \frac{7}{5} x$ $\displaystyle \frac{1}{5} = - \frac{7}{5} x$

$- \frac{5}{7} \cdot \frac{1}{5} = - \frac{5}{7}\left ( - \frac{7}{5} x \right )$ $\displaystyle - \frac{5}{7} \cdot \frac{1}{5} = - \frac{5}{7}\left ( - \frac{7}{5} x \right )$

$- \frac{1}{7} =x$ $\displaystyle - \frac{1}{7} =x$

The $\displaystyle x$-intercept is $\left (- \frac{1}{7} , 0 \right )$ $\displaystyle \left (- \frac{1}{7} , 0 \right )$.

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All Common Core: 8th Grade Math Resources

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Common Core: 8th Grade Math : Expressions & Equations

Study concepts, example questions & explanations for Common Core: 8th Grade Math

All Common Core: 8th Grade Math Resources

Example Questions

Example Question #1 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

Example Question #1 : How To Find X Or Y Intercept

Example Question #2 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

Example Question #2 : How To Find X Or Y Intercept

Example Question #3 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

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