Hannah is at the supermarket to buy carrots and potatoes. If she buys 3 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $10 do

[eddy2017]

Hannah is at the supermarket to buy carrots and potatoes. If she buys 3 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $10 dollars and if she buys 5 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $14 dollars. Which of these statements is correct:

a) carrots cost $2.50 a pound, which is more than potatoes cost.
b) carrots costs $2.00 a pound, which is the same as potatoes cost
c) carrots costs $ 2.00 a pound, which is less than potatoes cost.
d) potatoes cost $4.00 a pound, which is less than carrots cost.

I assigned variables to the cost of both carrots and to potatoes
let's say that carrots cost x dollars per pound
let's say that potatoes cost y dollars per pound.

If she buys 3 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $10
i can set this up in math notation like this:
3 * x + 4 * y=10

if she buys 5 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $14 dollars
5 * x + 4 * y =14

I have to equations here so I can solve it as a system of equations.
3x + 4y =10
5x +4y =14
I have studied that there are different methods to solve a system of linear equations : graphing. substitution method. elimination method.
If all this is correct up to here,
which is the method you would use, the best, the easiest, the fastest?
thank you in advance

 

[BigBeachBanana]

You can readily eliminate one of the variable?

 

[Deleted member 4993]

Hannah is at the supermarket to buy carrots and potatoes. If she buys 3 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $10 dollars and if she buys 5 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $14 dollars. Which of these statements is correct:

a) carrots cost $2.50 a pound, which is more than potatoes cost.
b) carrots costs $2.00 a pound, which is the same as potatoes cost
c) carrots costs $ 2.00 a pound, which is less than potatoes cost.
d) potatoes cost $4.00 a pound, which is less than carrots cost.

I assigned variables to the cost of both carrots and to potatoes
let's say that carrots cost x dollars per pound
let's say that potatoes cost y dollars per pound.

If she buys 3 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $10
i can set this up in math notation like this:
3 * x + 4 * y=10

if she buys 5 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $14 dollars
5 * x + 4 * y =14

I have to equations here so I can solve it as a system of equations.
3x + 4y =10
5x +4y =14
I have studied that there are different methods to solve a system of linear equations : graphing. substitution method. elimination method.
If all this is correct up to here,
which is the method you would use, the best, the easiest, the fastest?
thank you in advance

I would worry about being correct.

I would first find the correct answer and be done.

Some problem can have special angles, but I do not think it is worth wasting time to figure that "special" angle.

For this problem you may notice that the coefficients of 'y' s are already same so

[(5x +4y) - (3x+4y)] = 14 - 10 → 2x = 4 → x = 2 → y = 1 → does it make sense?

But that is "special" for this problem. Just keep your eyes open - and first time do it any "legitimate" way you can.

However, you can "judge" the statements a, b, c & d without calculating values of 'x' and 'y'.

 

[eddy2017]

You can readily eliminate one of the variable?

Yes, I think elimination works better. There is no variable with a coefficient of 1 so I am afraid I will run into Fractions if I go down the substitution road.
So,
3x + 4y = 10
5x + 4y = 14
I will find the MCM of 3 and 5 = 15
And as I will add both equations then it'll be convenient if I make one negative.
5 (3x + 4y =10)
-3(5x + 4y= 14)
15x + 20y = 50
-15x -12y= -42
Now I'll do the addition

8y=8
y=1
I'll choose one equation to replace y with its value
3x + 4y=10
3x +4(1)=10
3x +4=10
x = 2
So, now I have a coordinate point
(2,1)

 

[eddy2017]

I would worry about being correct.

I would first find the correct answer and be done.

Some problem can have special angles, but I do not think it is worth wasting time to figure that "special" angle.

For this problem you may notice that the coefficients of 'y' s are already same so

[(5x +4y) - (3x+4y)] = 14 - 10 → 2x = 4 → x = 2 → y = 1 → does it make sense?

But that is "special" for this problem. Just keep your eyes open - and first time d it any "legitimate" way you can.

However, you can "judge" the statements a, b, c & d without calculating values of 'x' and 'y'.

I'm interested in your statement, Dr Khan, and I will be pursuing it tomorrow, if you don't mind. It is kind of late now.
But it has been a good math day.
Thank you both bunches!

 

[JeffM]

If you must become comfortable with only one, choose substitution. It will work with systems that are not linear and gives exact answers.

 

[Steven G]

She buys 3 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $10 dollars and if she buys 5 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $14 dollars.

In the second case she is JUST buying an additional 2 pounds of carrots. The difference in costs is $4
$4 for two pounds of carrot means $2 per pound for carrots.

Now find the cost for potatoes.

 

[lev888]

Yes, I think elimination works better. There is no variable with a coefficient of 1 so I am afraid I will run into Fractions if I go down the substitution road.
So,
3x + 4y = 10
5x + 4y = 14
I will find the MCM of 3 and 5 = 15
And as I will add both equations then it'll be convenient if I make one negative.
5 (3x + 4y =10)
-3(5x + 4y= 14)
15x + 20y = 50
-15x -12y= -42
Now I'll do the addition

8y=8
y=1
I'll choose one equation to replace y with its value
3x + 4y=10
3x +4(1)=10
3x +4=10
x = 2
So, now I have a coordinate point
(2,1)

You wanted easy and fast. But then you did an extra step to eliminate x, while that step was not necessary to eliminate y.

 

[eddy2017]

You wanted easy and fast. But then you did an extra step to eliminate x, while that step was not necessary to eliminate y.

Yes, you're right lev. Just making sure the calculation was right. Until I get more used to solving these type of equations. The easiness and fastness was meant not so much for the amount of steps but for 'easy and fast to comprehend'.

 

[eddy2017]

She buys 3 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $10 dollars and if she buys 5 pounds of carrots and 4 pounds of potatoes, it will cost her a total of $14 dollars.

In the second case she is JUST buying an additional 2 pounds of carrots. The difference in costs is $4
$4 for two pounds of carrot means $2 per pound for carrots.

Now find the cost for potatoes.

3 pds of carrots × $2.00= $6.00
4 pds of potatoes × $1.00=4
6+4=10
In the second statement the cost of potatoes remains the same 1 pd cost $1.00.

 

[lev888]

Yes, you're right lev. Just making sure the calculation was right. Until I get more used to solving these type of equations. The easiness and fastness was meant not so much for the amount of steps but for 'easy and fast to comprehend'.

Well, if comprehension is your priority, I would say you need to know that the goal of this method is to eliminate variables one by one, doesn't matter in what order. In this case it's easier to eliminate y.

 

[eddy2017]

Well, if comprehension is your priority, I would say you need to know that the goal of this method is to eliminate variables one by one, doesn't matter in what order. In this case it's easier to eliminate y.

I know that. That is why the process is called 'elimination'. There is not much to it to be honest. But Jomo said I should become comfortable with the substitution method which the most exact and applicable in every situation. Do you agree?

 

[JeffM]

3 pds of carrots × $2.00= $6.00
4 pds of potatoes × $1.00=4
6+4=10
In the second statement the cost of potatoes remains the same 1 pd cost $1.00.

Eddy

Please do not take this the wrong way, but this thread is misconceived.

As tkhunny says, “right answers do not care how you found them.”

People who write text books on algebra usually try to use “nice” numbers in their problems. They are teaching algebra rather than arithmetic. if you ever want to use math for practical purposes, it is foolish to make judgments about algebraic methods based on the arithmetic simplicity of most teaching problems because most practical problems do not involve numbers carefully chosen to be nice.

Usually, elimination is easy IF the coefficients of the variables are small integers and the number of variables is small (as is true for this problem). In most other cases, elimination is messy and error prone.

As lev said, what is going on with all methods (except graphing) is systematic elimination of variables. That is the key point to understand. Substitution and elimination are just different technical ways to achieve the same goal.

Finally, I reiterate my earlier point: substitution is the most general method. If you have a problem involving four variables, how are you going to graph it?

 

[lev888]

I know that. That is why the process is called 'elimination'. There is not much to it to be honest. But Jomo said I should become comfortable with the substitution method which the most exact and applicable in every situation. Do you agree?

These are both "exact" methods. There are methods of solving equations that involve approximations (Numerical analysis). But in algebra all methods are exact.
Both methods are applicable to linear systems. You may choose one over the other depending on the system. E.g. If one equation already has one variable isolated it's a good candidate for substitution.
In different types of systems (non-linear) it may not be possible to use elimination, e.g. in one of the equations both variables are squared. Then use substitution, if possible.

 

[JeffM]

Based on lev’s immediately prior post, I see I need to clarify.

Substitution will work in cases where elimination will not always work, e.g. quadratic systems.

In systems of linear equations, both substitution and elimination will work and are exact.

Graphing will not work for systems of four or more variables and may not be exact even in a linear system of just two variables. . However, a graphing calculator may be the fastest way to get an approximate answer for a system in two variables.

 

[eddy2017]

Based on lev’s immediately prior post, I see I need to clarify.

Substitution will work in cases where elimination will not always work, e.g. quadratic systems.

In systems of linear equations, both substitution and elimination will work and are exact.

Graphing will not work for systems of four or more variables and may not be exact even in a linear system of just two variables. . However, a graphing calculator may be the fastest way to get an approximate answer for a system in two variables.

thank you so much. I think this post sums it up for me neatly. Thank you all!.