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3 years ago

Question: -3x – 10 = 32

Mathematics

2 answers:

zlopas [31]3 years ago

8 0

Answer:

x=−14

Step-by-step explanation:

8_murik_8 [283]3 years ago

4 0

Answer:

x = -14

Step-by-step explanation:

The first thing you are trying to do is get the x alone. so first you will look at 10 and do the opposite of the sign that you have. So you would take the -10 and make it +10. you would solve -10 + 10 and 32 + 10 because whatever you do to one side you do to the other. -10 +10= 0 and 32+10 = 42. Your equation now is -3x=42. we do not put -3x+0=42 because zero does not have a value. From here we would look at the -3x and do the opposite sign, which means dividing. whenever a number is next to x or any variable it is multiplication. For example, 7x is the same as 7*x. The opposite of multiplication is division. so we would do -3x/-3. We do 42/-3 because whatever we do to one side we do to the other. On the left side we are left with x then you add the equal sign and the quotient of 42/-3.

Answer: x = -14

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3 years ago

Which equation has solutions of 6 and -6? x2 – 12x + 36 = 0 x2 + 12x – 36 = 0 x2 + 36 = 0 x2 – 36 = 0

Svetlanka [38]

<u>Answer:</u>

The equation that has solutions 6 and -6 is $x^2 - 36 = 0$

<u>Solution:</u>

We have to find which equation has the solutions 6 and -6.

We have been given three equations.

$x^{2}-12 x+36=0$ --- eqn 1

$x^{2}+12 x-36=0$ -- eqn 2

$x^{2}-36=0$ ---- eqn 3

The 6 and -6 to satisfy any of these equations they have to be the roots of the equation.

This means that when we substitute 6 and -6 in any of the equations and then solve them the answer on simplification should be 0.

This condition should individually be satisfied by both 6 and -6 for any one of the equations.

Now let us try and substitute 6 and -6 in eq1.

Now, substituting 6 in eq1.

62-12×6+36=0

Now we simply the equation to check is the LHS is equal to the RHS of the equation.

LHS:

72-72=0

RHS: 0

Since LHS=RHS it is the root of the equation.

Now we check if -6 satisfies eq1.

-62-12×-6+36=0

LHS:

72+72=144

RHS: 0

Hence LHS is not equal to RHS, -6 is not the root of eq1.

Similarly we check for eq2

Checking for 6 and -6 we get

LHS is not equal to RHS hence this does not satisfy eq2.

Now in the same way we check for eq3

LHS=RHS for both 6 and -6 hence they are the solutions for eq3.

Hence the equation that has solutions 6 and -6 is $x^2 - 36 = 0$

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(5+W)(W+4)

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8 0

3 years ago

Please help. Thank you! :)

Tcecarenko [31]

Hello!

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