Tag Archives: algebra 1

11.07.14
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Slope

This week’s topic is:  Distance, Midpoint, and Slope.

Q:  The slope of the line passing through (-1,-3) and (7,y) is -1/2.  What is the value of y?

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Answer: -7

Algebraic solution by hand:

Use the slope formula:

 

2014-11-07_153414

Solution using my TI-84 Plus program DISTANCE:

Run DISTANCE.  How?

Use your multiple choice answer choices.  One of them is -7.  (If you don’t have choices, draw a picture.  Plot the point (-1,-3) and do a slope of -1/2 from there – down 1, right 2, down 1, right 2, … until you get to a point where x is 7.  Read the y-value of that point – it will be -7 and check it using this program.)

x1=-1

y1=-3

x2=7

y2=-7

Ignore the distance information – we don’t need it for this problem.  Press ENTER for the midpoint information, which we don’t need either.  Press ENTER once more to find that the slope is -1/2.

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11.06.14
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Distance

This week’s topic is:  Distance, Midpoint, and Slope.

Q:  For what value(s) of x is the point (x,4) exactly 5 units away from the point (6,8)?

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Answer: 3 or 9

Solution by hand:

Use the distance formula:

distance

Solution using my TI-84 Plus program DISTANCE:

Draw a picture:

distgraph

Guess that x=2, maybe.  It’s probably an integer if this is an SAT problem.  Use the answer choices if available.

Run DISTANCE.  How?

x1=2

y1=4

x2=6

y2=8

The distance is 5.66.  It’s too big – you wanted it to be 5 – so move the point a little closer to the middle.

Maybe x=3:

x1=3

y1=4

x2=6

y2=8

The distance is exactly 5.  Perfect.  (There’s also another possible answer, x=9.  If you have multiple choice answers, you will see 3 and 9 as a choice, so try both and see that they both work.  For a grid-in SAT problem, you would only have to find one answer anyway.)           :

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10.30.14
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Ratios

Q: Last month, the ratio of sunny days to rainy days was 3:2.  How many rainy days  were there last month, if the month had 30 days?  (Assume all days were either sunny or rainy.)

 

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Answer: 12

Solution by hand:

Number of sunny days = 3x

Number of rainy days = 2x

3x + 2x = 30

5x=30

x = 6

There were 2x rainy days, so there were 12 rainy days.

 

Solution using my TI-84 Plus Program RATIO:

Run RATIO.  How?

Total number of items (days) = 30

Number of categories of items = 2  (sunny and rainy)

Ratio part for category 1 (sunny) = 3

Ratio part for category 2 (rainy) = 2

Distribution is: sunny = 18 and rainy = 12.

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10.29.14
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Proportions

Q:  x is inversely proportional to w.  If x = 6 when w = 8, what is x when w = 12?

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Answer: 4

Solution by hand:

For inverse proportions, (x1)(w1) = (x2)(w2)

(6)(8) = (x)(12)

48 = 12x

x=4

Solution using my TI-84 Plus Program PROPORTN:

Run PROPORTN.  How?

x1 = 6

y1 = 8  (here, we have renamed w1 as y1)

Have y2 (really w2)

y2 = 12

Choose ‘Inverse’ proportion.

Then, x2 = 4.

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10.27.14
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Ratios

Q:  The ratio of blue cars to non-blue cars in the parking lot is 2:7.  If there are 1260 cars in the parking lot, how many are blue?

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Answer: 280

Solution by hand:

Number of blue cars = 2x

Number of non-blue cars = 7x

2x + 7x = 1260

9x=1260

x = 140

There are 2x blue cars, so there are 280 blue cars.

 

Solution using my TI-84 Plus Program RATIO:

Run RATIO.  How?

Total number of items = 1260

Number of categories of items = 2  (blue and non-blue)

Ratio part for category 1 (blue) = 2

Ratio part for category 2 (non-blue) = 7

Distribution is: blue = 280 and non-blue = 980.

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09.12.14
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ACT , Algebra 1 – system of equations

MathPro Q&A Forum (ACT)

 Q: For what value of A will the following system of equations have infinitely many solutions?

3x – 2y = 4

Ax + 4y = -8

 

Answer: -6

Explanation: In order for the system to have infinitely many solutions, the two equations must describe the same line.  (Recall that the solution(s) of a system are the points where the graphs of the two lines meet.  For the two lines to meet in infinitely many points, they must be the same line.)

 To make the second equation the same as the first, multiply both sides of the first equation by -2:

 -2(3x – 2y) = -2(4)

-6x + 4y = -8

So, A = -6.

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09.10.14
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Algebra 1 – Find the equation of a line

Q: If the slope of the line through (-3,6) and (1,y) is 1/2, what is the value of y?

Answer:  8

Explanation: Slope is computed by the formula (y2 – y1) / (x2 – x1) … the change in y divided by the change in x.

So,

(y – 6) / (1 – -3) = 1 / 2

(y – 6) / 4 = 1 / 2

Multiply both sides by 4 and you have:

y – 6 = 2

y = 8

BUT, if you’re taking a test and having trouble remembering the method above, try this instead: Plot the point (-3,6) on a graph and move from that point with a slope of 1/2. Move up 1, right 2, up1, right 2… and repeat this pattern until your x-value is 1. Then, read the y-value at the same point. This is not my favorite way, but it works!

And one more thing!  If this were a multiple choice question, you could use my TI-84 Plus program ‘DISTANCE’ (which finds distance, midpoint, and slope for any two ordered pairs).  Substitute one of the answer choices for x, and see if the program reports that slope is 1/2.

 

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09.10.14
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Algebra 2 – proportions

Q:  If x is inversely proportional to the square of y, and x = 4 when y = 1, find x when y = 2.

Answer:  1

Explanation:  We are told that x and y² are inversely proportional, so x • y² = k, where k is a constant.

Let’s find out what k is, using the given info:

4 • 1² = k

k = 4

Now,  x • y² = 4.

Plug in the new value for y, to find the new value for x:

x • y² = 4

x • 2² = 4

x • 4 = 4

x = 1

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09.10.14
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Algebra 1, 2 – systems of equations

Q: Red pens cost twice as much as blue pens. Two blue pens and three red pens cost a total of $2.32. How much does a blue pen cost?

 Answer:  $0.29

 Explanation:  Let r be the cost of a red pen, and let b be the cost of a blue pen. 

Then, r = 2b and 2b + 3r = 2.32. Solve this system of equations. 

By substitution,

r + 3r = 2.32

4r = 2.32

r = 0.58

Plug r= 0.58 into the first equation (or second, but I like the first better):

0.58 = 2b

0.29 = b

 You can also use other algebraic methods such as elimination, Cramer’s Rule, etc.  Or use my TI-84 Plus program CRAMER once you rearrange the system to say: 2b – r = 0 and 2b + 3r = 2.32.

 

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09.10.14
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SAT – Counting and Ordering

Q:  20 students are to choose a president and vice president from among themselves. How many possible ways can they be chosen?

Answer:  380

Explanation 1:  Using the Counting Principle, set up two blanks to represent the president and vice president, from left to right:

____    ____

There are 20 choices for the president.  Put 20 in the first blank:

_20_    ____

The same person cannot be chosen for both positions, so there are 19 left to choose from for the vice president.  Put 19 in the second blank:

_20_    _19_

Now, multiply these numbers.  20 x 19 = 380.

Explanation 2:  This is a permutation problem since you are choosing 2 people from a total of 20 people, and the order is important.  The order is important because the two people chosen must be given specific positions, and switching those positions results in a different scenario.  Do  20 nPr 2 , or use the formula for nPr.  On the TI-84 Plus, find nPr in MATH-PRB-2.  The result is 380.

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