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3 years ago

How do you solve equations that contain multiplication or division

2 answers:

6 0

You multiply or divide, depending on what is contained in the equation. But if there is addition or subtraction in the equation as well, then you do that too. But the order you do it in all depends on how the equation is written, what you're solving for, whether there are parentheses or not, and if there are, where they are placed.
If there's a particular problem you're stuck on that involves multiplication/division, feel free to post it or pm me about it and I'll try to help out.

kodGreya [7K]3 years ago

5 0

Always perform the same operation to both sides of the equation. When you multiply or divide, you have to multiply and divide by the entire side of the equation. Try to perform addition and subtraction first to get some multiple of x by itself on one side.

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ALGEBRAIC EXPRESSION 11. Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and 2x + 4y – 7. 12.

Naily [24]

Answer:

Explained below.

Step-by-step explanation:

(11)

Subtract the sum of (13x - 4y + 7z) and (- 6z + 6x + 3y) from the sum of (6x - 4y - 4z) and (2x + 4y - 7z).

$[(6x - 4y - 4z) +(2x + 4y - 7z)]-[(13x - 4y + 7z) + (- 6z + 6x + 3y) ]\\=[6x-4y-4z+2x+4y-7z]-[13x-4y+7z-6z+6x+3y]\\=6x-4y-4z+2x+4y-7z-13x+4y-7z+6z-6x-3y\\=(6x+2x-13x-6x)+(4y-4y+4y-3y)-(4z+7z+7z-6z)\\=-11x+y-12z$

Thus, the final expression is (-11x + y - 12z).

(12)

From the sum of (x² + 3y² - 6xy), (2x² - y² + 8xy), (y² + 8) and (x² - 3xy) subtract (-3x² + 4y² - xy + x - y + 3).

$[(x^{2} + 3y^{2} - 6xy)+(2x^{2} - y^{2} + 8xy)+(y^{2} + 8)+(x^{2} - 3xy)] - [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[x^{2} + 3y^{2} - 6xy+2x^{2} - y^{2} + 8xy+y^{2} + 8+x^{2} - 3xy]- [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[4x^{2}+3y^{2}-xy+8]-[-3x^{2} + 4y^{2} - xy + x - y + 3]\\=4x^{2}+3y^{2}-xy+8+3x^{2}-4y^{2}+xy-x+y-3\\=7x^{2}-y^{2}-x+y+5$

Thus, the final expression is (7x² - y² - x + y + 5).

(13)

What should be subtracted from (x² – xy + y² – x + y + 3) to obtain (-x²+ 3y²- 4xy + 1)?

$A=(x^{2} - xy + y^{2} - x + y + 3) - (-x^{2}+ 3y^{2}- 4xy + 1)\\=x^{2} - xy + y^{2} - x + y + 3 +x^{2}- 3y^{2}+ 4xy -1\\=2x^{2}-2y^{2}+3xy-x+y+2$

Thus, the expression is (2x² - 2y² + 3xy - x + y + 2).

(14)

What should be added to (xy – 3yz + 4zx) to get (4xy – 3zx + 4yz + 7)?

$A=(4xy-3zx + 4yz + 7)-(xy - 3yz + 4zx) \\=4xy-3zx + 4yz + 7 -xy + 3yz - 4zx\\=3xy-7zx+7yz+7$

Thus, the expression is (3xy - 7zx + 7yz + 7).

(15)

How much is (x² − 2xy + 3y²) less than (2x² − 3y² + xy)?

$A=(2x^{2} - 3y^{2} + xy)-(x^{2} - 2xy + 3y^{2})\\=2x^{2} - 3y^{2} + xy-x^{2} + 2xy - 3y^{2}\\=x^{2}-6y^{2}+3xy$

Thus, the expression is (x² - 6y² + 3xy).

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3 years ago

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Nat2105 [25]

Vertical angles congruence theorem, then SAS

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3 years ago

What is the five-number summary for this data set?

Natali [406]

Answer:

Step-by-step explanation:

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3 years ago

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kompoz [17]

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3 years ago

Using triangle A B C as the pre-image and the origin as the center of dilation, which coordinate point is not a coordinate of a

Ostrovityanka [42]

Answer:

The correct option is:

begin ordered pair 4 comma 12 end ordered pair i.e. (4,12).

Step-by-step explanation:

We are given the coordinates of the point A,B and C of the triangle ΔABC as:

A(-1,3) , B(2,1) and C(-2,-1).

Now it is given that there is dilation of this triangle about the origin with a scale factor of 4.

This means that the change in the coordinates after the transformation is given by the rule:

(x,y) → (4x,4y)

As A → A' , B → B' and C → C'

Hence the coordinates A',B' and C' are given by:

A' are (4×-1,4×3)=(-4,12)

B' are (4×2,4×1)=(8,4)

and C' are (4×-2,4×-1)=(-8,-4).

Hence the coordinate point which is not a coordinate of dilation is:

begin ordered pair 4 comma 12 end ordered pair i.e. (4,12)

3 0

3 years ago