I need help with this problem
Solve the following system of equations using elimination by multiplication.

If the given equations are indeed

[tex]\begin{cases}x-\frac4{5y} = 9\frac45 \\ -14x - 4y = -46\end{cases}[/tex]

we can solve by substitution. Solve the second equation for x :

[tex]-14x - 4y = -46[/tex]

[tex]\implies 7x + 2y = 23[/tex]

[tex]\implies7x = 23 - 2y[/tex]

[tex]\implies x = (23 - 2y)/7[/tex]

Substitute this into the first equation and solve for y :

[tex]\dfrac{23-2y}7 - \dfrac4{5y} = 9\dfrac45[/tex]

On the right side, write the mixed number as an improper fraction:

[tex]9 + \dfrac45 = \dfrac{45}5 + \dfrac45 = \dfrac{45+4}5 = \dfrac{49}5[/tex]

[tex]\implies \dfrac{23-2y}7 - \dfrac4{5y} = \dfrac{49}5[/tex]

Multiply both sides of this equation by 35y (the LCM of the denominator of all three fractions)

[tex]35y \times \dfrac{23-2y}7 - 35y \times \dfrac4{5y} = 35y \times \dfrac{49}5[/tex]

[tex]5y\times(23-2y) - 7\times4 = 7y \times 49[/tex]

[tex]115y - 10y^2 - 28 = 343y[/tex]

[tex]10y^2 + 228y + 28 = 0[/tex]

[tex]5y^2 + 114y + 14 = 0[/tex]

To solve the quadratic, I'll complete the square :

[tex]5\left(y^2 + \dfrac{114}5y\right) + 14 = 0[/tex]

[tex]5\left(\left(y + \dfrac{57}5\right)^2 - \left(\dfrac{57}5\right)^2 \right) + 14 = 0[/tex]

[tex]5\left(y + \dfrac{57}5\right)^2 - \dfrac{57^2}5 + 14 = 0[/tex]

[tex]5\left(y + \dfrac{57}5\right)^2 = -\dfrac{3179}5[/tex]

This system has no real solutions since the square of any real number must be positive.

If we allow complex numbers, we can continue solving to end up with two complex solutions for y,

[tex]\left(y + \dfrac{57}5\right)^2 = -\dfrac{3179}{25}[/tex]

[tex]y + \dfrac{57}5 = \pm i \sqrt{\dfrac{3179}{25}}[/tex]

[tex]y = -\dfrac{57}5 \pm i\dfrac{17\sqrt{11}}5[/tex]

and we can go on to solve for x.

Hence my comment; I suspect you meant to write the system

[tex]\begin{cases}x-\frac45 y = 9\frac45 \\ -14x - 4y = -46\end{cases}[/tex]

because its solution is far simpler. Multiplying through the first equation by 5 gives

[tex]5\times x - 5 \times \dfrac45 y = 5 \times 9\dfrac45[/tex]

[tex]\implies 5x - 4y = 49[/tex]

(since we know 9 + 4/5 = 49/5)

Meanwhile, multiplying through the second equation by -1 gives

[tex]-1 \times (-14x) + (-1) \times (-4y) = -1 \times (-46)[/tex]

[tex]\implies 14x + 4y = 46[/tex]

So if we combine the two equations, we can eliminate y and solve for x :

[tex](5x - 4y) + (14x + 4y) = 49 + 46[/tex]

[tex]\implies 19x = 95[/tex]

[tex]\implies \boxed{x = 5}[/tex]

and solving for y gives

[tex]14x + 4y = 46[/tex]

[tex]\implies 70 + 4y = 46[/tex]

[tex]\implies 4y = -24[/tex]

[tex]\implies \boxed{y = -6}[/tex]

(10²)²

(EXPONENT FORM ONLY)

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