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2 years ago

Solve by using substitution . SHOW YOUR WORK. 3х + y = –17 бx — бу = –18

Mathematics

2 answers:

lilavasa [31]2 years ago

4 0

Step-by-step explanation:

1. 3x+y+17=0

2.x-y+3=0

Sorry mam I use mental math

weqwewe [10]2 years ago

4 0

Answer: X = —5 and Y = —2

Step-by-step explanation: What we have is a pair of simultaneous equations;

3X + Y = —17 —————-(1)

6X — 6Y = —18 ————(2)

We shall use the substitution method as instructed. Looking at equation (1),

We shall make Y the subject of the equation, hence

Y = —17 —3X

We shall now substitute for the value of Y into equation (2)

6X — 6Y = —18

6X — 6(—17 —3X) = —18

6X + 102 + 18X = —18

By collecting like terms we now have

24X = —18 —102

(Note that when a positive value crosses to the other side of the equation it becomes negative and vice versa)

24X = —120

Divide both sides of the equation by 24

X = —5

Having calculated the value of X as —5, we can now calculate Y as follows

From equation (1)

3X + Y = —17

3(—5) + Y = —17

—15 + Y = —17

Add 15 to both sides of the equation

—15 + 15 + Y = —17 + 15

Y = —2

Therefore, X = —5 and Y = —2

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Ksju [112]

Answer:

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Step-by-step explanation:

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3 years ago

Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

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

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