Choose the equivalent system of linear equations that will produce the same solution as the one given below. 4x - y = -11 2x + 3y = 5

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Choose the equivalent system of linear equations that will produce the same solution as the one given below. 4x - y = -11 2x + 3y = 5

Mathematics
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Have you solved systems of linear equations before? If so, do you have any ideas of where to start?
no i have idea
no idea**

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4x - y = -11 2x + 3y = 5 Here's the steps to these kind of questions. Step 1: Choose one of the equation above and solve for one variable x or y (depends on your choice) Step 2: After you solve for, let's say X, you plug whatever you got for x into the second equation that you didn't began with. Step 3: When you plug in what you solved for x into the second equation, all the variables will become y. You solve for y, and you'll get a number. Plug that number to either one of the equations and see what you'll get for x. If none of that seems familiar with you, then i will do my best to help and make you understand this.
Step 1: Which equation are you choosing from: ? 4x - y = -11 2x + 3y = 5
does it matter which one i choose
Nope. It does not matter.
Which equation is the easiest to solve for y?
I'd go with the easiest and i'll choose the first equation.
2x+3y=5 there is no negatives so its probably easier to solve
Okay, we'll go with that.
oops lol we can go with yours
Okay.
i'll choose the variable y because it's alone by itself. We can get rid of negatives when we divide both sides by negative :)
\(\large 4x - y = -11 \) Subtract both sides by 4x \(\large 4x-4x - y = -11-4x \) \(\large - y = -11-4x \) Divide both sides by -1 \(\Large \frac{- y}{-1} = \frac{-11-4x}{-1} \) \(\large y = 11+4x \)
Makes sense?
what is y equal to now?
14x ? im sorry im so confused on this, and my teacher isnt responding
Why 14 x? We can't add 11 with 4x because they are different terms. Y is equal to 11+4x, that's the only info we know
idk what to do
What does step 2 say?
After you solve for, let's say X, you plug whatever you got for x into the second equation that you didn't began with.
Yes. But let's not deal with x since we are dealing with y. So, in different wording, we take whatever we got for y and plug it for the y in the second equation. Makes sense?
y=11+4x Our second equation is 2x+3y=5 \(\large 2x+3(11+4x)=5 \)
2x+33+12x=5 ? or do you not use distributive property
Yes, you already distributed it
yup, you're right so far, now let's keep going right now we have: 2x+33+12x=5 now add the x's together: 2x + 12x = __x fill in the blank
how do i solve that? add or multiply?
so 14x
oh yeah sorry
yes, so now we have 14x + 33 = 5 subtract 33 from each sign
*from each side
-19=-28
am i supposed to subtract 33 from 33 too or no
not quite, remember that you can't combine variables and numbers 14x + 33 - 33 = 14x 5 - 33 = -28 14x = -28, make sense so far?
yes
14x = -28 divide both sides by 14
x=-2
good, we're almost done we know that x = -2 now, so we can find y let's go back to our first equation: 4x - y = -11 can you tell me what y is now?
-2
not quite. let's try plugging in -2 into the equation 4*(-2) - y = -11 -8 - y = -11 now can you tell me what y is?
no i dont understand
-8 - y = -11 add 8 to both sides, what do you get?
-y=-19
add, not subtract -11 + 8 = ?
-3
right, so -y = -3 what is y?
3
boom, we're done x = -2 and y = 3 our solution is (-2,3)
thanks
Solve the system of equations by substitution. What is the solution for x? 2x + y = 1 4x + 2y = -2 can you help with this one too?

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