Solve Find the value of a and b whether (3 - sqrt(5))/(3 + 2sqrt(5)) = a * sqrt(5) - 19/11 * b | 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

$$\frac { 3 - \sqrt { 5 } } { 3 + 2 \sqrt { 5 } } = a \sqrt { 5 } - \frac { 19 } { 11 } b$$

Solve for a

$a=\frac{19\sqrt{5}b}{55}-\frac{19\sqrt{5}}{55}+\frac{9}{11}$

Steps for Solving Linear Equation

Rationalize the denominator of $\frac{3-\sqrt{5}}{3+2\sqrt{5}}$ by multiplying numerator and denominator by $3-2\sqrt{5}$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{\left(3+2\sqrt{5}\right)\left(3-2\sqrt{5}\right)}=a\sqrt{5}-\frac{19}{11}b$$

Consider $\left(3+2\sqrt{5}\right)\left(3-2\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{3^{2}-\left(2\sqrt{5}\right)^{2}}=a\sqrt{5}-\frac{19}{11}b$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-\left(2\sqrt{5}\right)^{2}}=a\sqrt{5}-\frac{19}{11}b$$

Expand $\left(2\sqrt{5}\right)^{2}$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-2^{2}\left(\sqrt{5}\right)^{2}}=a\sqrt{5}-\frac{19}{11}b$$

Calculate $2$ to the power of $2$ and get $4$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\left(\sqrt{5}\right)^{2}}=a\sqrt{5}-\frac{19}{11}b$$

The square of $\sqrt{5}$ is $5$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\times 5}=a\sqrt{5}-\frac{19}{11}b$$

Multiply $4$ and $5$ to get $20$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-20}=a\sqrt{5}-\frac{19}{11}b$$

Subtract $20$ from $9$ to get $-11$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{-11}=a\sqrt{5}-\frac{19}{11}b$$

Use the distributive property to multiply $3-\sqrt{5}$ by $3-2\sqrt{5}$ and combine like terms.

$$\frac{9-9\sqrt{5}+2\left(\sqrt{5}\right)^{2}}{-11}=a\sqrt{5}-\frac{19}{11}b$$

The square of $\sqrt{5}$ is $5$.

$$\frac{9-9\sqrt{5}+2\times 5}{-11}=a\sqrt{5}-\frac{19}{11}b$$

Multiply $2$ and $5$ to get $10$.

$$\frac{9-9\sqrt{5}+10}{-11}=a\sqrt{5}-\frac{19}{11}b$$

Add $9$ and $10$ to get $19$.

$$\frac{19-9\sqrt{5}}{-11}=a\sqrt{5}-\frac{19}{11}b$$

Multiply both numerator and denominator by -1.

$$\frac{-19+9\sqrt{5}}{11}=a\sqrt{5}-\frac{19}{11}b$$

Divide each term of $-19+9\sqrt{5}$ by $11$ to get $-\frac{19}{11}+\frac{9}{11}\sqrt{5}$.

$$-\frac{19}{11}+\frac{9}{11}\sqrt{5}=a\sqrt{5}-\frac{19}{11}b$$

Swap sides so that all variable terms are on the left hand side.

$$a\sqrt{5}-\frac{19}{11}b=-\frac{19}{11}+\frac{9}{11}\sqrt{5}$$

Add $\frac{19}{11}b$ to both sides.

$$a\sqrt{5}=-\frac{19}{11}+\frac{9}{11}\sqrt{5}+\frac{19}{11}b$$

The equation is in standard form.

$$\sqrt{5}a=\frac{19b+9\sqrt{5}-19}{11}$$

Divide both sides by $\sqrt{5}$.

$$\frac{\sqrt{5}a}{\sqrt{5}}=\frac{19b+9\sqrt{5}-19}{11\sqrt{5}}$$

Dividing by $\sqrt{5}$ undoes the multiplication by $\sqrt{5}$.

$$a=\frac{19b+9\sqrt{5}-19}{11\sqrt{5}}$$

Divide $\frac{9\sqrt{5}+19b-19}{11}$ by $\sqrt{5}$.

$$a=\frac{19\sqrt{5}b}{55}-\frac{19\sqrt{5}}{55}+\frac{9}{11}$$

Solve for b

$b=\frac{\sqrt{5}\left(11a-9\right)+19}{19}$

Steps for Solving Linear Equation

Rationalize the denominator of $\frac{3-\sqrt{5}}{3+2\sqrt{5}}$ by multiplying numerator and denominator by $3-2\sqrt{5}$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{\left(3+2\sqrt{5}\right)\left(3-2\sqrt{5}\right)}=a\sqrt{5}-\frac{19}{11}b$$

Consider $\left(3+2\sqrt{5}\right)\left(3-2\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{3^{2}-\left(2\sqrt{5}\right)^{2}}=a\sqrt{5}-\frac{19}{11}b$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-\left(2\sqrt{5}\right)^{2}}=a\sqrt{5}-\frac{19}{11}b$$

Expand $\left(2\sqrt{5}\right)^{2}$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-2^{2}\left(\sqrt{5}\right)^{2}}=a\sqrt{5}-\frac{19}{11}b$$

Calculate $2$ to the power of $2$ and get $4$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\left(\sqrt{5}\right)^{2}}=a\sqrt{5}-\frac{19}{11}b$$

The square of $\sqrt{5}$ is $5$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\times 5}=a\sqrt{5}-\frac{19}{11}b$$

Multiply $4$ and $5$ to get $20$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-20}=a\sqrt{5}-\frac{19}{11}b$$

Subtract $20$ from $9$ to get $-11$.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{-11}=a\sqrt{5}-\frac{19}{11}b$$

Use the distributive property to multiply $3-\sqrt{5}$ by $3-2\sqrt{5}$ and combine like terms.

$$\frac{9-9\sqrt{5}+2\left(\sqrt{5}\right)^{2}}{-11}=a\sqrt{5}-\frac{19}{11}b$$

The square of $\sqrt{5}$ is $5$.

$$\frac{9-9\sqrt{5}+2\times 5}{-11}=a\sqrt{5}-\frac{19}{11}b$$

Multiply $2$ and $5$ to get $10$.

$$\frac{9-9\sqrt{5}+10}{-11}=a\sqrt{5}-\frac{19}{11}b$$

Add $9$ and $10$ to get $19$.

$$\frac{19-9\sqrt{5}}{-11}=a\sqrt{5}-\frac{19}{11}b$$

Multiply both numerator and denominator by -1.

$$\frac{-19+9\sqrt{5}}{11}=a\sqrt{5}-\frac{19}{11}b$$

Divide each term of $-19+9\sqrt{5}$ by $11$ to get $-\frac{19}{11}+\frac{9}{11}\sqrt{5}$.

$$-\frac{19}{11}+\frac{9}{11}\sqrt{5}=a\sqrt{5}-\frac{19}{11}b$$

Swap sides so that all variable terms are on the left hand side.

$$a\sqrt{5}-\frac{19}{11}b=-\frac{19}{11}+\frac{9}{11}\sqrt{5}$$

Subtract $a\sqrt{5}$ from both sides.

$$-\frac{19}{11}b=-\frac{19}{11}+\frac{9}{11}\sqrt{5}-a\sqrt{5}$$

Reorder the terms.

$$-\frac{19}{11}b=-\sqrt{5}a+\frac{9}{11}\sqrt{5}-\frac{19}{11}$$

The equation is in standard form.

$$-\frac{19}{11}b=-\sqrt{5}a+\frac{9\sqrt{5}}{11}-\frac{19}{11}$$

Divide both sides of the equation by $-\frac{19}{11}$, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{19}{11}b}{-\frac{19}{11}}=\frac{-\sqrt{5}a+\frac{9\sqrt{5}}{11}-\frac{19}{11}}{-\frac{19}{11}}$$

Dividing by $-\frac{19}{11}$ undoes the multiplication by $-\frac{19}{11}$.

$$b=\frac{-\sqrt{5}a+\frac{9\sqrt{5}}{11}-\frac{19}{11}}{-\frac{19}{11}}$$

Divide $\frac{9\sqrt{5}}{11}-\sqrt{5}a-\frac{19}{11}$ by $-\frac{19}{11}$ by multiplying $\frac{9\sqrt{5}}{11}-\sqrt{5}a-\frac{19}{11}$ by the reciprocal of $-\frac{19}{11}$.

$$b=\frac{11\sqrt{5}a}{19}-\frac{9\sqrt{5}}{19}+1$$