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Kobotan [32]
3 years ago
7

What is 4(4m-3)-(m-5)=52 step by step

Mathematics
1 answer:
ch4aika [34]3 years ago
8 0

Answer:

-3

Step-by-step explanation:

Simplifying

4(4m + -3) + -1(m + -5) = -52

Reorder the terms:

4(-3 + 4m) + -1(m + -5) = -52

(-3 * 4 + 4m * 4) + -1(m + -5) = -52

(-12 + 16m) + -1(m + -5) = -52

Reorder the terms:

-12 + 16m + -1(-5 + m) = -52

-12 + 16m + (-5 * -1 + m * -1) = -52

-12 + 16m + (5 + -1m) = -52

Reorder the terms:

-12 + 5 + 16m + -1m = -52

Combine like terms: -12 + 5 = -7

-7 + 16m + -1m = -52

Combine like terms: 16m + -1m = 15m

-7 + 15m = -52

Solving

-7 + 15m = -52

Solving for variable 'm'.

Move all terms containing m to the left, all other terms to the right.

Add '7' to each side of the equation.

-7 + 7 + 15m = -52 + 7

Combine like terms: -7 + 7 = 0

0 + 15m = -52 + 7

15m = -52 + 7

Combine like terms: -52 + 7 = -45

15m = -45

Divide each side by '15'.

m = -3

Simplifying

m = -3

Hope this helped :)

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Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

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2. Solve for x

\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}

\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}

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3 years ago
How do i do this, im really confused and idk how to do it at this point
nalin [4]

Answer:

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Step-by-step explanation:

7 0
3 years ago
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