Question Video: Using the Properties of Parallel Lines and Solving Linear Equations to Determine the Value of an Unknown | Nagwa Question Video: Using the Properties of Parallel Lines and Solving Linear Equations to Determine the Value of an Unknown | Nagwa

# Question Video: Using the Properties of Parallel Lines and Solving Linear Equations to Determine the Value of an Unknown Mathematics • First Year of Secondary School

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In the figure, π΄π΅ = 3π₯, π΅πΆ = 5π₯, π·πΈ = (3π₯ β 6), and πΈπΉ = (4π₯ β 3). Find the value of π₯.

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### Video Transcript

In the figure, π΄π΅ is equal to three π₯, π΅πΆ is equal to five π₯, π·πΈ is equal to three π₯ minus six, and πΈπΉ is equal to four π₯ minus three. Find the value of π₯.

The first thing we can do is go ahead and label our figure. π΄π΅ equals three π₯. π΅πΆ equals five π₯. π·πΈ equals three π₯ minus six. And πΈπΉ equals four π₯ minus three. Now what we see happening in this figure is three parallel lines are cutting two transversals. And when this happens, the transversals are cut proportionally. And that means π΄π΅ over π·πΈ will be proportional to π΅πΆ over πΈπΉ. So what we can do is plug in the values we know for each of these four segments so that we have three π₯ over three π₯ minus six is equal to five π₯ over four π₯ minus three. Be careful here; we canβt cancel out the three π₯βs since the denominator is three π₯ minus six.

Since we canβt do any simplification, in order to solve, weβll want to cross multiply, which means weβll have three π₯ times four π₯ minus three is equal to five π₯ times three π₯ minus six. We can distribute, and then weβll have 12π₯ squared minus nine π₯ is equal to 15π₯ squared minus 30π₯. We wanna get all the π₯-values on the same side, so we subtract 12π₯ squared from both sides and add nine π₯ to both sides. On the left, it only leaves zero. And on the right, weβll have three π₯ squared minus 21π₯.

From there, thereβs a factor of three that we can remove. If we divide both sides of the equation by three, weβll see that zero equals π₯ squared minus seven π₯. And then, we wanna undistribute this factor of π₯. π₯ squared minus seven π₯ can be rewritten as π₯ times π₯ minus seven. And then weβll take each of these terms and set them equal to zero. Weβll have π₯ equals zero and π₯ minus seven equals zero. This means the options for π₯ here are π₯ equals zero or π₯ equals seven.

Since weβre dealing with distance, we know that π₯ equals zero is not a valid solution, which leaves us with π₯ equals seven. If you wanted to, you could go back and see what each line segment measured. π΄π΅ would be 21, π΅πΆ would be 35, π·πΈ would be 15, and πΈπΉ would be 25. And then we could check, is 21 over 15 equal to 35 over 25? 21 divided by 15 is 1.4. 35 divided by 25 is also 1.4, which means these are correct proportions, and π₯ must be equal to seven.

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