Question

Subtract 4xy²z – x²y + xyz² from 2xyz² – xy²z + x²y.

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Answer 1

Answer:

1. Substitution

x-2y = 9

-3x+y = -7

Substitution works by solving for one of the variables in one of the equations. When you solve for this variable, you'll still have the other variable to deal with, so this would not be the answer for the variable. I'll explain this more once we get there. So, let's solve for x in the first equation:

x-2y = 9

x = 2y+9 (add 2y to each side of the equation).

As you can see, we still have the y to deal with. Therefore, the answer "x = 2y + 9" will not work. This is only a step to actually solve for x and see what its value is as a number. Since we have solved for x in this way, we need to plug this in the other equation (not the first equation) to solve for y:

-3x+y = -7

-3(2y+9)+y = -7 (substituting x = 2y+9 into the equation)

-6y-27+y = -7 (distributive property or multiplying -3 and 2y and -3 and 9)

-5y-27 = -7 (combine like terms)

-5y = 20 (add 27 to each side of the equation)

y = -4 (divide by -5 on each side of the equation).

If you tried solving for y in the equation you rearranged for x = 2y+9, you wouldn't get anywhere. That is why we must plug that into the other equation. We have found y to equal -4. Now we must plug -4 in for y in either equation to find x. I'll do both:

x-2y = 9

x-2(-4) = 9 (substituting y = -4)

x+8 = 9 (multiplying -2 and -4)

x = 1 (subtracting 8 from each side);

-3x+y = -7

-3x+(-4) = -7 (substituting -4 in for y)

-3x = -3 (adding 4 to each side of the equation)

x = 1 (dividing each side by -3).

As you can see, no matter which equation we plug y = -4 into, we get the same value for x. But this is usually not acceptable unless the problem states to only find x or y. We usually need to write this as a pair of coordinates or a point. So, with x first and y second, the point would be (1, -4) in the Cartesian coordinate system (rectangular coordinate system). To check ourselves, we just plug 1 in for x and -4 in for y in both equations and see if they are true:

x-2y = 9

1-2(-4) = 9 (substituting x = 1 and y = -4)

1+8 = 9 (multiplying -2 and -4)

9 = 9 (combining like terms)

Step-by-step explanation:

-3(1)+(-4) = -7 (substituting x = 1 and y = -4)

-3 + (-4) = -7 (multiplying -3 and 1)

-7 = -7 (combining like terms)

As you can see, both equations check out and the point (1, -4) is the solution to this system.

2. Elimination

x-2y = 9

-3x+y = -7

This method manipulates one of the equations in such a way that we can add (or subtract) a variable by the ability of the polynomial operation of addition (or subtraction). In other words, we can manipulate one of the equations in a way to "eliminate" a variable. This way we can solve for the variable that hasn't been eliminated. Let us eliminate the x by manipulating the first equation. Here's my work:

x-2y = 9

-3x+y = -7

We need the first x to equal the same number (regardless of sign) as the second x. We can make the first x equal 3 or -3, and I'll show both ways of doing so. Doing the positive first, we need to make x go to 3x. To do this, all we need to do is multiply the entire first equation by 3 (3 * x = 3x, which is what we need):

3(x-2y = 9)

3x-6y = 27 (multiplying each term by 3).

Now we can eliminate by adding the like terms vertically:

3x-6y = 27 (using the rewritten equation)

-3x+y = -7

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

0x-5y = 20.

Since 0x = 0 (any number multiplied by 0 is 0), we are left with -5y = 20, which we find y to equal -4. Let us now change the first equation so that x has a -3 by it:

-3(x-2y = 9)

-3x+6y = -27 (multiply each term by -3)

And elimination:

-3x+6y = -27

-3x+y = -7

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

0x+5y = -20 [if we do -3, we have to subtract each term because -3x-(-3x) = -3x +3x = 0x, which is what we need].

We could have eliminated the y's first and solved for x. Either way, we would come up with the answer (1, -4) as in substitution. For both methods we must be wary of signs.

Answer 2

${\rm{\underline{Subtract}}}:-$

$\rm1.) {4xy}^{2}z - {x}^{2} y + {xyz}^{2} \: \: \: from \: 2xyz^{2} - {xy}^{2} z + {x}^{2} y$

$\rm \longmapsto \bigg( 2xyz^{2} - {xy}^{2} z + {x}^{2} y \bigg) - \bigg({4xy}^{2}z - {x}^{2} y + {xyz}^{2} \bigg)$

$\rm \longmapsto 2xyz^{2} - {xy}^{2} z + {x}^{2} y - {4xy}^{2}z + {x}^{2} y - {xyz}^{2}$

Now we will have to pair them according to the alphabets.

$\rm \longmapsto 2xyz^{2}- {xyz}^{2} - {xy}^{2} z- {4xy}^{2}z + {x}^{2} y + {x}^{2} y$

$\rm \longmapsto {xyz}^{2} - {5xy}^{2}z + 2 {x}^{2} y$

${\boxed{ \underline{ \underline{ \longmapsto {\rm{ \red {{xyz}^{2} - {5xy}^{2}z + 2 {x}^{2} y}}}}}}} \red{\bigstar}$