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Comment on System Question
Would you mind explaining, in
2x+y+z = 3
5x+3y+z= 8
Here we don't start by subtraction. WHY?!
Our goal with systems of
In almost all cases (the above practice question excluded from this, since we aren't asked to determine the individual values of x and y), our goal with systems of equations is to reduce the number of variables we're working with. To do that, we need the same coefficient in both equations.
Consider the following system:
3x + 5y = 13
3x + y = 5
Since the both x terms already have the same coefficient (3), we can simply subtract the bottom equation from the top equation to get a new (much nicer!) equation: 4y = 8, which we can then solve to get y = 2. And so on.
Now consider this system:
2x + 5y = 17
x + y = 4
In this case, the two equations don't share any of the same coefficients. So, subtracting one equation from the other isn't going to help us.
To see what I mean, let's subtract the bottom equation from the top equation. When we do this we get a new equation: x + 4y = 13.
Now what? We really can't do anything with this new equation. We certainly can't solve it for x or y.
In order for us to eliminate one of the variables (and create a new equation with only 1 variable), we must first take one of the original equations (2x + 5y = 17 and x + y = 4) and do something with it in order to transform that equation into an equivalent equation that shares a coefficient with the other equation.
Let's do that.
We have:
2x + 5y = 17
x + y = 4
One option is to take the bottom equation and multiply both sides of the equation by 2 to get an EQUIVALENT equation: 2x + 2y = 8.
So, our system now looks like this:
2x + 5y = 17
2x + 2y = 8
Now let's subtract the bottom equation from the top equation. When we do this we get a new equation: 3y = 9.
PERFECT! We can solve this to get y = 3, and we can use this fact to solve for x and get x = 1. DONE!
But there's more than one way to get the same coefficients in both equations. Here's another approach using the same system:
2x + 5y = 17
x + y = 4
In this new approach, let's take the bottom equation and multiply both sides of the equation by 5 to get an EQUIVALENT equation: 5x + 5y = 20.
So, our system now looks like this:
2x + 5y = 17
5x + 5y = 20
Now let's subtract the bottom equation from the top equation. When we do this we get a new equation: -3x = -3.
PERFECT! We can solve this to get x = 1, and we can use this fact to solve for y and get y = 3. DONE!
Does that help? You might also want to review the video on solving systems of equations at https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...
Given 2x + y + z = 3 and 5x +
A. -2
B. -1
C. 0
D. 1
E. 2
Given: 2x + y + z = 3 and 5x + 3y + z = 8
Take 2x + y + z = 3 and multiply both sides by 2 to get the EQUIVALENT equation: 4x + 2y + 2z = 6
We now have:
5x + 3y + z = 8
4x + 2y + 2z = 6
This question where did that 2 come from? and tell me this above method shown is not elimination method ? as we multiply both the equations in that method. When we just multiply one equation with a certain constant does it not cause chaos or destabilize the whole question?
Okay, let's first answer your
Okay, let's first answer your question about whether multiplying both sides of an equation destabilizes the equation/system.
When we multiply both sides of an equation by some value, we create an EQUIVALENT equation.
Consider the equation: x + y = 5
If we multiply both sides by 2, we get an EQUIVALENT equation: 2x + 2y = 10. Notice that this equation doesn't tell us anything new about x and y. The original equation (x+y = 5) tells us that the sum of x and y is 5. The "new" EQUIVALENT equation tells us that if we first double x and y, then their sum is 10.
Likewise, consider this very simple equation: x = 3
If we multiply both sides by 2, we get the EQUIVALENT equation: 2x = 6. Nothing new here.
So, multiplying both sides of an equation by a number just creates an equivalent equation.
In a way it's no different than creating equivalent fractions.
For example, we can take 1/2 and multiply top and bottom by 5 to get an equivalent fraction 5/10. Just as rewriting 1/2 as 5/10 does not destabilize anything, so rewriting x + y = 5 as 2x + 2y = 10 does not destabilize anything either.
Now let's go to this new question you have posed.
First notice that the question is NOT asking us to solve the system for the individual values of x, y and z.
Instead, we are asked to determine the value of x + y - z?
So, is there some way we can manipulate the given equations to create a new equation that will tell us the value of x + y - z?
Well, we're given this:
2x + y + z = 3
5x + 3y + z = 8
If we subtract the bottom equation from the top equation, we get: -3x - 2y = -5
Does this help us determine the value of x+y-z? No. Okay, what else can we do?
2x + y + z = 3
5x + 3y + z = 8
If we ADD the two equations, we get: 7x + 4y + 2z = 11
Does this help us determine the value of x+y-z? No. Okay, what else can we do?
Perhaps we can create some EQUIVALENT equations to help us out.
At this point, we should recognize that, since we're trying to find the value of 1x + 1y - 1z, we want to create equivalent equations such that when one equation is subtracted from the other, the coefficients become 1, 1 and -1.
At this point, we should start playing around with different scenarios in our head.
After a while, we might recognize that, if we take 2x + y + z = 3 and multiply both sides by 2, we get the EQUIVALENT equation: 4x + 2y + 2z = 6
So, our two equations are:
5x + 3y + z = 8
4x + 2y + 2z = 6
At this point, when we subtract the bottom equation from the top equation, we get: x + y - z = 2
VOILA!
Thank you so much! :)
Can you please explain how 36
Also since this equation involves fractions, I found it easier to find the LCM of 3x and 2x, which is 6. Which left me with -7Y = 9. So Y = 9/7.
And I found the LCM of -4y and -5y, which is 20. Which left me with -7X=5.
So X=-5/7.
Then once I added them i got -14/7 which equals -2.
3 - 36/7 = 3/1 - 36/7
3 - 36/7 = 3/1 - 36/7
= 21/7 - 36/7
= (21 - 36)/7
= -15/7
Your LCM approach is good too.
Of course, the big goal here is to reduce our calculations.
So, recognizing that we can subtract the second equation from the first equation to determine the answer is key to saving us a TON of time.
I am a bit confused because I
Please see;
3x-4y = 3
2x-5y = 5
From the 1st equation
x = 3+4y/3
Substitute into 2x -5y =5
2(3 + 4y/3) - 5y = 5
= (6+8y/3) -5y = 5
= 6 + 8y -5y = 15
= 6 +3y = 15
3y = 15-6
y= 3
Now substitute y =3 into equation 1
3x -4(3)=3
3x -12 =3
3x = 15
x= 5
Hence, x+y = 5 + 3 = 8
P.s I know that I am wrong but I just cant tell where I got it wrong. Please help thank you
The mistake occurs between
The mistake occurs between these two lines of your solution:
= (6 + 8y)/3 - 5y = 5
= 6 + 8y - 5y = 15
We're multiplying both sides by 3, so the second line should be:
= 6 + 8y - 15y = 15 (15y not 5y)
Cheers,
Brent