What this lesson covers
This one runs the same equation twice, once with choices and once without, and lands on the part students skip: the choices were doing your checking for you, so on a grid-in you substitute your own answer back in before you move on.
00:00 Nothing to test
00:14 The safety net
00:29 The method
01:05 Three ways this goes wrong
01:22 Recap
The equation and its choices are original, written for SAT Climb.
Worked examples
The questions the video works, written out: the setup, each step, the answer and the trap.
0:29Example 1
What is the value of x? The video shows the same question twice: once with the choices A) 7 B) 11 C) 17 D) 21, and once with no choices at all.
- With choices, you never need the algebra. Test each value in both sides. A, 7: 3(3) = 9 and 2(7) + 5 = 19, no. B, 11: 3(7) = 21 and 2(11) + 5 = 27, no. C, 17: 3(13) = 39 and 2(17) + 5 = 39, yes. D, 21: 3(17) = 51 and 2(21) + 5 = 47, no.
- Without choices, that whole strategy is gone. Solve it. Distribute the 3: .
- Take 2x off both sides: . Add 12 to both sides: x = 17.
- The choices used to be your check, so now you are. Substitute back: 3(17 - 4) = 3(13) = 39, and 2(17) + 5 = 34 + 5 = 39. Both sides match.
- Type 17. Two characters, no x, no equals sign.
Answer: x = 17, typed as 17
Backsolving needs choices to solve backwards from; a grid-in has none, so the algebra is the only way in. The lesson is not which of 7, 11, 17 and 21 is right. It is that skipping the check is the trap: nothing on the screen is there to disagree with you, so substituting back is the only thing standing between right work and a wrong entry. And a blank costs exactly what a wrong entry costs, so always type something.
One more, same method
What is the value of x?
- No choices, so solve. Distribute the 4: .
- Take 4x off both sides: . Add 9 to both sides: . Divide by 3: x = 7.
- Check: 4(7 + 3) = 4(10) = 40, and 7(7) - 9 = 49 - 9 = 40. Both sides match.
- Type 7.
Answer: x = 7, typed as 7
Solve, check, type: the check replaces the answer choices. A student who distributes the 4 over x only, writing 4x + 3, reaches and x = 4, and with no choices on the screen nothing flags it. Substituting 4 back gives 28 on the left and 19 on the right, which is how that slip gets caught.
Lesson transcript
The narration of the video, word for word, under its chapter headings.
With four choices you always have a way in. You can test them. One of the four has to work, so you can reach the answer without ever doing the algebra. Here is the same question twice. On the left, with choices. Try seventeen. Three times thirteen is thirty nine, and two times seventeen plus five is thirty nine. Done, and you never touched the algebra. On the right, the choices are gone, and so is that whole strategy. You have to solve it. Three x minus twelve equals two x plus five, so x is seventeen. Then do the job the choices used to do for you. Put it back in and check it. Three ways this goes wrong. Hunting for a strategy that needs choices. Skipping the check, because nothing on the screen is there to disagree with you. And leaving it blank, when a wrong entry costs exactly the same. No choices. Solve it, then check it.