Solve انشر العبارات التالية A = 4x(2x + 3); B = 6x(A + 3) + 2ax(3 + 1) أحسب قيمة A من أجل بطريقتين مختلفتين ؟ | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\left. \begin{array} { l } { 4 x ( 8 x + 3 ) } \\ { 6 x ( x ( x + 3 ) + 2 ) + 2 x \times ( 5 + 4 ) } \\ { 6 x ( x + 3 ) + 2 x \times ( 5 + 4 ) } \\ { 9 x - 120 + 5 x = 2 20 } \\ { \text { 3 \right.$$

Solve for x, y, z, a

$y = \frac{211520}{49} = 4316\frac{36}{49} \approx 4316.734693878$
$z = \frac{3996000}{343} = 11650\frac{50}{343} \approx 11650.145772595$
$a = \frac{58560}{49} = 1195\frac{5}{49} \approx 1195.102040816$

Solution Steps

Consider the first equation. Combine $9x$ and $5x$ to get $14x$.

$$14x-120=2\times 20$$

Multiply $2$ and $20$ to get $40$.

$$14x-120=40$$

Add $120$ to both sides.

$$14x=40+120$$

Add $40$ and $120$ to get $160$.

$$14x=160$$

Divide both sides by $14$.

$$x=\frac{160}{14}$$

Reduce the fraction $\frac{160}{14}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{80}{7}$$

Consider the second equation. Insert the known values of variables into the equation.

$$y=4\times \frac{80}{7}\left(8\times \frac{80}{7}+3\right)$$

Multiply $4$ and $\frac{80}{7}$ to get $\frac{320}{7}$.

$$y=\frac{320}{7}\left(8\times \frac{80}{7}+3\right)$$

Multiply $8$ and $\frac{80}{7}$ to get $\frac{640}{7}$.

$$y=\frac{320}{7}\left(\frac{640}{7}+3\right)$$

Add $\frac{640}{7}$ and $3$ to get $\frac{661}{7}$.

$$y=\frac{320}{7}\times \frac{661}{7}$$

Multiply $\frac{320}{7}$ and $\frac{661}{7}$ to get $\frac{211520}{49}$.

$$y=\frac{211520}{49}$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=6\times \frac{80}{7}\left(\frac{80}{7}\left(\frac{80}{7}+3\right)+2\right)+2\times \frac{80}{7}\left(5+4\right)$$

Multiply $6$ and $\frac{80}{7}$ to get $\frac{480}{7}$.

$$z=\frac{480}{7}\left(\frac{80}{7}\left(\frac{80}{7}+3\right)+2\right)+2\times \frac{80}{7}\left(5+4\right)$$

Add $\frac{80}{7}$ and $3$ to get $\frac{101}{7}$.

$$z=\frac{480}{7}\left(\frac{80}{7}\times \frac{101}{7}+2\right)+2\times \frac{80}{7}\left(5+4\right)$$

Multiply $\frac{80}{7}$ and $\frac{101}{7}$ to get $\frac{8080}{49}$.

$$z=\frac{480}{7}\left(\frac{8080}{49}+2\right)+2\times \frac{80}{7}\left(5+4\right)$$

Add $\frac{8080}{49}$ and $2$ to get $\frac{8178}{49}$.

$$z=\frac{480}{7}\times \frac{8178}{49}+2\times \frac{80}{7}\left(5+4\right)$$

Multiply $\frac{480}{7}$ and $\frac{8178}{49}$ to get $\frac{3925440}{343}$.

$$z=\frac{3925440}{343}+2\times \frac{80}{7}\left(5+4\right)$$

Multiply $2$ and $\frac{80}{7}$ to get $\frac{160}{7}$.

$$z=\frac{3925440}{343}+\frac{160}{7}\left(5+4\right)$$

Add $5$ and $4$ to get $9$.

$$z=\frac{3925440}{343}+\frac{160}{7}\times 9$$

Multiply $\frac{160}{7}$ and $9$ to get $\frac{1440}{7}$.

$$z=\frac{3925440}{343}+\frac{1440}{7}$$

Add $\frac{3925440}{343}$ and $\frac{1440}{7}$ to get $\frac{3996000}{343}$.

$$z=\frac{3996000}{343}$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$a=6\times \frac{80}{7}\left(\frac{80}{7}+3\right)+2\times \frac{80}{7}\left(5+4\right)$$

Multiply $6$ and $\frac{80}{7}$ to get $\frac{480}{7}$.

$$a=\frac{480}{7}\left(\frac{80}{7}+3\right)+2\times \frac{80}{7}\left(5+4\right)$$

Add $\frac{80}{7}$ and $3$ to get $\frac{101}{7}$.

$$a=\frac{480}{7}\times \frac{101}{7}+2\times \frac{80}{7}\left(5+4\right)$$

Multiply $\frac{480}{7}$ and $\frac{101}{7}$ to get $\frac{48480}{49}$.

$$a=\frac{48480}{49}+2\times \frac{80}{7}\left(5+4\right)$$

Multiply $2$ and $\frac{80}{7}$ to get $\frac{160}{7}$.

$$a=\frac{48480}{49}+\frac{160}{7}\left(5+4\right)$$

Add $5$ and $4$ to get $9$.

$$a=\frac{48480}{49}+\frac{160}{7}\times 9$$

Multiply $\frac{160}{7}$ and $9$ to get $\frac{1440}{7}$.

$$a=\frac{48480}{49}+\frac{1440}{7}$$

Add $\frac{48480}{49}$ and $\frac{1440}{7}$ to get $\frac{58560}{49}$.

$$a=\frac{58560}{49}$$

The system is now solved.

$$x=\frac{80}{7}$$ $$y=\frac{211520}{49}$$ $$z=\frac{3996000}{343}$$ $$a=\frac{58560}{49}$$