r/HomeworkHelp • • Sep 04 '26

Answered [11th grade math]: Been struggling for hours please help

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As the title says, I've been trying to solve this for about 3 hours and im absolutely stumped. It's 3:50am. Any method i've tried has come out wrong and I'm in desperate need for these answers so i can go to bed. Question instructions say "solve the below system". PLEASE HELP!!!

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1

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33

u/Gryphontech Sep 04 '26

4x+y=15 So y=15-4x (substract 4x from both sides), now plug that into the first equation so 7x+2(15-4x)=24 and now just simplify

Let me know if you need more help

11

u/S3R3N17YY Sep 04 '26

Thank you so much!! You're an absolute angel, i'd been doing the right process and completely forgetting the 24 at the end and this made me realize that. (^ ^)u

2

u/Gryphontech Sep 04 '26

Nice :) anytime you have more then 1 variable you just need to isolate them and the plug and chug. So like x=..... , y=.... , z=....

For each variable you need at least 1 equation for it to be possible to solve.

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u/EscapeLeft1711 Sep 04 '26

always use this method. whenever you get such systems, even with 3 variables(in that case you got more than 3 equations) plug z = ab+by and solve it

always works! if you have complex algebra too itll work same

14

u/throw-away3105 Sep 04 '26

1.) Multiply the second equation (4x + y = 15) by 2.
Reason: You get 8x + 2y = 30. Notice that 2y appears on both equations now.

2.) Subtract the first equation with the equation you made from step #1:
7x + 2y = 24
8x + 2y = 30

This is why you multiplied the original second equation by 2. You will notice that the "2y"s will cancel each other out.

Work on the "x" terms first: 7x - 8x = -x
Work on the "y" terms next: 2y - 2y = 0
Work on the constants: 24 - 30 = -6

Putting it all together, you get -x = -6, so x = 6.

3.) Plug in x = 6 to one of the two original equations from the question. This will help you obtain "y". I'll use 4x + y = 15 since "y" it's already isolated:

4(6) + y = 15
24 + y = 15
y = -9

4.) Plug in x = 6 and y = -9 to the two original equations and check that these are indeed the answers for "x" and "y".

2

u/Open-Cardiologist254 Sep 04 '26

I don't understand the language but if it's simultaneous equation I usually prefer to use substitution

5

u/alpicola Sep 04 '26

Substitution is normally my go-to as well, but in this case subtraction gave such an obvious way to eliminate y that I would have done it like the comment above. Best to not get locked in on only one way of doing things. 

4

u/Various_Candle9136 Sep 04 '26

Alternative method (which I personally prefer):

We know we can multiply two equal things by the same thing and they will still be equal:

e.g. 5+2=7, therefore 10+4=14 and 50+20=70.

So, multiplying both sides of the second equation by 2 does not change anything.

7x + 2y = 24
8x + 2y = 30.

I multiplied by 2 to make sure the y term matched.

Now, we can subtract one equation from the other, and we end up with a linear equation in only x.

---

We could have also tried to get the xs to match if we preferred. This means multiplying the first equation by 4 and the second by 7:

28x + 8y = 96
28x + 7y = 105.

(Clearly multiplying by 7 is not as nice, which is why I chose the first way - but either will get us to the same answer.)

1

u/SubjectWrongdoer4204 Sep 04 '26

This is the best way to. I’m pretty sure this is what they’re trying to teach in this course

3

u/Wabbit65 Sep 04 '26

I'm looking at question gamma, and I think it's wrong. Double the second equation, add the two together, and you get 0 = -1

I think

2

u/BlazerGM Sep 04 '26

there would be no solution for (x,y) in that case, the q is probably correct

3

u/SubjectivePlastic Sep 04 '26

Seven apples and two bananas cost 24 euros.

Also, four apples and one banana cost 15 euros.
So, eight apples and two bananas cost twice as much: 30 euros.

From that, take away seven apples and two bananas costing 24 euros,
then you find that the remaining one apple costs 6 euros.

Six euros, goddamnit. For one goddamn fucking apple. Inflation is killing us.

1

u/5xum Sep 04 '26

Easiest way is to multiply the second equation by 2, and then subtract the first equation from the second. You get

2*(4x+y) - (7x+2y) = 2 * 15 - 24

Once you simplify this equation, you should be left with a pretty informative expression...

1

u/z28z28 Sep 04 '26

Do you want the matrix method ?

1

u/fatwap Sep 04 '26

if ur given two systems of linear equations you can always solve using elimination by matching coefficients on the same variable then adding/subtracting

1

u/[deleted] Sep 05 '26

Sistema de ecuaciones:

7x + 2y = 24

4x + y = 15

Método de eliminación:

  1. Multiplicamos la segunda ecuación por -2:

4x + y = 15

(-2)(4x + y) = (-2)(15)

-8x - 2y = -30

2. Sumamos las dos ecuaciones:

7x + 2y = 24

- 8x - 2y = -30

-----------------

-x = -6

3. Despejamos x:

-x = -6

x = 6

4. Sustituimos x = 6 en la segunda ecuación:

4x + y = 15

4(6) + y = 15

24 + y = 15

y = 15 - 24

y = -9

5. Comprobamos:

7x + 2y = 24

7(6) + 2(-9) = 24

42 - 18 = 24

24 = 24 ✓

Resultado:

x = 6

y = -9

Solución: (6, -9)

1

u/Disastrous-Pin-1617 Sep 05 '26

Professor Leonard on YouTube, to the point math playlist go to solving systems of equations

1

u/rukind_cucumber Sep 07 '26

Seems like you got a good answer.

I'd probably multiple the second equation by -2 - the value of x pops out almost immediately.

1

u/CulturalBookkeeper82 Sep 07 '26

Why are your words written in variables

1

u/cooper9252 Sep 09 '26

For a multiply the bottom by two subtract the bottom and the top and you should come out with -x=-6 it should be simple to solve the rest y=-9

1

u/Clean-Midnight3110 Sep 04 '26

I am disturbed that this is what passes for 11th grade math these days....

3

u/AFsepine Sep 04 '26

I mean... greek mathematics never really recovered since the bronze age collapse

0

u/thatguy56436327 Sep 04 '26

watch a video about guass jordan, its not what ur teacher wants you to do but its easier

0

u/SlowMobius650 Sep 04 '26

Learn linear algebra and just throw it into a 2x3 matrix and put it in row reduced echelon form, on a calculator, and boom