Need help to even begin with

[dollyayesha2345]

Show that the equations $$8a-10b=\frac{1}{3}$$ and $$20a-25b=\frac{5}{6}$$ have many solutions.

Please tell me how to even begin with and I'll solve and share my inputs

 

[Otis]

Please tell me how to even begin

Hi Dolly. Are you familiar with Elimination or Substitution methods, for solving a system of two equations? Those are two methods taught in beginning algebra.

When we use either of those approaches, sometimes the variables end up disappearing -- leaving us with a single equation that is obviously true, like 5=5.

Whenever that happens, it indicates that the system has infinite solutions. Graphically speaking, it means that the two given equations both graph as the same line (i.e., the equations are equivalent).

There are hundreds of lessons and worked examples online. Google keywords elimination method. Let us know, if you see anything you don't understand. Otherwise, please study the method steps, and then try your exercise. Show us what you've done, if you get stuck.

TIP: You may clear the two fractions, before starting. Multiply each equation by the fraction's denominator.

 

[dollyayesha2345]

Hi Dolly. Are you familiar with Elimination or Substitution methods, for solving a system of two equations? Those are what's taught in beginning algebra.

When we use either of those approaches, sometimes the variables end up disappearing -- leaving us with a single equation that is obviously true, like 5=5.

Whenever that happens, it indicates that the system has infinite solutions. Graphically speaking, it means that the two given equations both graph as the same line (i.e., the equations are equivalent).

There are hundreds of lessons and worked examples online. Google keywords elimination method. Let us know, if you see anything you don't understand.

So I have to use the Elimination method right? I'll solve and share my answer asap.
I guess the category I posted is wrong I believe.

 

[Otis]

So I have to use the Elimination method

You may use either elimination or substitution.

We may have cross-posted, as I had just added some information to my first reply. Did you see the tip?

 

[dollyayesha2345]

You may use either elimination or substitution.

We may have cross-posted, as I had just added some information to my first reply. Did you see the tip?

Yes yes indeed. I'll just solve and show asap.

 

[dollyayesha2345]

You may use either elimination or substitution.

We may have cross-posted, as I had just added some information to my first reply. Did you see the tip?

is this correct? if yes, then what do I do after this.. how do I prove whether the equation has many solutions

 

[Otis]

You copied the second equation incorrectly. The fraction is 5/6.

The elimination method involves eliminating one of the variables. Did you study any of the worked examples?

Before adding (or subtracting) the two equations, you first need to adjust coefficients for one of the variables.

You have:

24a - 30b = 1

120a - 150b = 5

Now, if the coefficients on variable a were opposite numbers, then adding the equations would eliminate variable a (because opposites add to zero).

But, those two coefficients are not opposites. They are 24 and 120.

What number could you multiply the first equation by, to change 24a into -120a?

Do that, and then add the equations.

 

[dollyayesha2345]

nope

You copied the second equation incorrectly. The fraction is 5/6.

The elimination method involves eliminating one of the variables. Did you study any of the worked examples?

Before adding (or subtracting) the two equations, you first need to adjust coefficients for one of the variables.

You have:

24a - 30b = 1

120a - 150b = 5

Now, if the coefficients on variable a were opposite numbers, then adding the equations would eliminate variable a (because opposites add to zero).

But, those two coefficients are not opposites. They are 24 and 120.

What number could you multiply the first equation by, to change 24a into -120a?

Do that, and then add the equations.

nopes, I didn't go through any of the worked examples.

 

[Otis]

You chose to obtain 120a in both equations, instead of my suggestion to get opposites 120a and -120a. That's okay, but you can't eliminate by adding 120a+120a. When the coefficients are the same, then you eliminate by subtracting equations (because 120-120 gives zero).

By the way, you didn't add your b terms correctly (-150 plus -150 is -300).

So, after multiplying the first equation by 5, did you notice that your two equations were the same? Subtracting one from the other yields 0=0.

That indicates infinite solutions.

You need a lot more practice. Study some lessons, and work through the examples. That's how we learn.

 

[dollyayesha2345]

You chose to obtain 120a in both equations, instead of my suggestion to get opposites 120a and -120a. That's okay, but you can't eliminate by adding 120a+120a. When the coefficients are the same, then you eliminate by subtracting equations (because 120-120 gives zero).

By the way, you didn't add your b terms correctly (-150 plus -150 is -300).

So, after multiplying the first equation by 5, did you notice that your two equations were the same? Subtracting one from the other yields 0=0.

That indicates infinite solutions.

You need a lot more practice. Study some lessons, and work through the examples. That's how we learn.

thank you so much! in school we were taught that the sign of second equations shouldn't be considered like if its + we automatically change it to - and so on.
You were the one who told me to add the equations after multiplying. nonetheless thank you so much once again!

 

[Otis]

we were taught that the sign of second equations shouldn't be considered

That's wrong. When using the elimination method, it is VERY important to consider the signs of terms in BOTH equations.

You were the one who told me to add the equations after multiplying.

Yes, but I'd also instructed you to change 24a into -120a. You didn't follow that instruction. You'd changed 24a to +120a, instead. THAT is why we needed to switch to subtraction. Please slow down and pay closer attention.

You do not understand the process. I was willing to guide you on this exercise, but brief guidance is not a good way to learn a brand new topic. (There's a lot of important, related information that I haven't included in this thread.) You need to study actual, detailed lessons. You need to work through several examples, step by step, until your brain has been exposed to the patterns enough for you to comprehend and remember.

Do you have difficulty understanding your textbook? If so, you may google the topic. There are hundreds of alternative lessons available. Try to find some that you can understand. Cheers

?

 

[Steven G]

To confirm that there are many solutions, which means that there are infinitely many solutions, I would decide what to multiply 1/3 by to get 5/6. Then I would multiply the 1st equation (the one with 1/3) by this magic number and confirm that both equations now are identical.

 

[Otis]

I would multiply the 1st equation ... by this magic number and confirm that both equations now are identical.

Yup. Similar to showing that both simplify to the same equation (4a-5b=1/6).

Too bad the OP wasn't able to tell us what their class has been doing. Maybe it's linear algebra.

?

 

[dollyayesha2345]

Yup. Similar to showing that both simplify to the same equation (4a-5b=1/6).

Too bad the OP wasn't able to tell us what their class has been doing. Maybe it's linear algebra.

?

I am not in any class so how should I answer this! all I have is the question(s) at my end.

 

[Steven G]

Figure out what to multiply 1/3 by to get 5/6. Then I would multiply the 1st equation (the one with 1/3) by this magic number and confirm that both equations now are identical.

Hint: 1 times what equals 5 and 3 times what equals 6.

 

[Otis]

I am not in any class so how should I answer this! all I have is the question(s) at my end.

Hi Dolly. You hadn't provided any context, so we assumed you were a student (please see 'Read Before Posting').

The answer to how you "should" do something depends on context (like, why you're interested). From where did you get the exercise, and why? What is your math background?

There are different ways to answer, so you have potential choices! Depending on your situation, you may be free to pick from any one or more approaches. I'd given two common, classroom suggestions: elimination or substitution. You could also change the equations into slope intercept form. You could do as Jomo suggested, multiplying to match constant terms. You could also show that both equations are equivalent to (multiples of) a third equation. You could plot both equations on the same graph, to show the lines coincide. Any demonstration that shows the given equations as equivalent is an answer. There are even more, specialized approaches (beyond secondary school math).

 

[dollyayesha2345]

Figure out what to multiply 1/3 by to get 5/6. Then I would multiply the 1st equation (the one with 1/3) by this magic number and confirm that both equations now are identical.

Hint: 1 times what equals 5 and 3 times what equals 6.

The magic numbers are 5 & 2

 

[JeffM]

Dolly

Jomo said magic number, not numbers. The number he meant is written $\dfrac{5}{2}$.

 

[dollyayesha2345]

up u

Dolly

Jomo said magic number, not numbers. The number he meant is written $\dfrac{5}{2}$.

I was legit today years old to know that fractions are called magic numbers ?

 

[Otis]

I was legit today years old to know that fractions are called magic numbers

That English is not clear, Dolly, but Jomo did not say that fractions are called magic numbers. He was referring to a case-specific factor that changes one equation to match another. In other words, when Jomo said 'magic' he meant 'special'. Just like 5 was a special factor, when it changed 24a-30b=1 to match the other equation 120a-150b=5.

Neither do I have the textbook nor am I in any class.

Why are you interested in this exercise?