Solve (7+sqrt(5))/(7-sqrt(5))-(7-sqrt(5))/(7+sqrt(5))=x+7sqrt(5)y | 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

$$\frac{ 7+ \sqrt{ 5 } }{ 7- \sqrt{ 5 } } - \frac{ 7- \sqrt{ 5 } }{ 7+ \sqrt{ 5 } } =x+7 \sqrt{ 5 } y$$

Solve for x

$x=\frac{7\sqrt{5}\left(1-11y\right)}{11}$

Short Solution Steps

Rationalize the denominator of $\frac{7+\sqrt{5}}{7-\sqrt{5}}$ by multiplying numerator and denominator by $7+\sqrt{5}$.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Consider $\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Square $7$. Square $\sqrt{5}$.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{49-5}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Subtract $5$ from $49$ to get $44$.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Multiply $7+\sqrt{5}$ and $7+\sqrt{5}$ to get $\left(7+\sqrt{5}\right)^{2}$.

$$\frac{\left(7+\sqrt{5}\right)^{2}}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(7+\sqrt{5}\right)^{2}$.

$$\frac{49+14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

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

$$\frac{49+14\sqrt{5}+5}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Add $49$ and $5$ to get $54$.

$$\frac{54+14\sqrt{5}}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Rationalize the denominator of $\frac{7-\sqrt{5}}{7+\sqrt{5}}$ by multiplying numerator and denominator by $7-\sqrt{5}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}=x+7\sqrt{5}y$$

Consider $\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}=x+7\sqrt{5}y$$

Square $7$. Square $\sqrt{5}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{49-5}=x+7\sqrt{5}y$$

Subtract $5$ from $49$ to get $44$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{44}=x+7\sqrt{5}y$$

Multiply $7-\sqrt{5}$ and $7-\sqrt{5}$ to get $\left(7-\sqrt{5}\right)^{2}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)^{2}}{44}=x+7\sqrt{5}y$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(7-\sqrt{5}\right)^{2}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{49-14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}=x+7\sqrt{5}y$$

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

$$\frac{54+14\sqrt{5}}{44}-\frac{49-14\sqrt{5}+5}{44}=x+7\sqrt{5}y$$

Add $49$ and $5$ to get $54$.

$$\frac{54+14\sqrt{5}}{44}-\frac{54-14\sqrt{5}}{44}=x+7\sqrt{5}y$$

Since $\frac{54+14\sqrt{5}}{44}$ and $\frac{54-14\sqrt{5}}{44}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{54+14\sqrt{5}-\left(54-14\sqrt{5}\right)}{44}=x+7\sqrt{5}y$$

Do the multiplications in $54+14\sqrt{5}-\left(54-14\sqrt{5}\right)$.

$$\frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}=x+7\sqrt{5}y$$

Do the calculations in $54+14\sqrt{5}-54+14\sqrt{5}$.

$$\frac{28\sqrt{5}}{44}=x+7\sqrt{5}y$$

Divide $28\sqrt{5}$ by $44$ to get $\frac{7}{11}\sqrt{5}$.

$$\frac{7}{11}\sqrt{5}=x+7\sqrt{5}y$$

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

$$x+7\sqrt{5}y=\frac{7}{11}\sqrt{5}$$

Subtract $7\sqrt{5}y$ from both sides.

$$x=\frac{7}{11}\sqrt{5}-7\sqrt{5}y$$

Solve for y

$y=-\frac{\sqrt{5}x}{35}+\frac{1}{11}$

Steps for Solving Linear Equation

Rationalize the denominator of $\frac{7+\sqrt{5}}{7-\sqrt{5}}$ by multiplying numerator and denominator by $7+\sqrt{5}$.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Consider $\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Square $7$. Square $\sqrt{5}$.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{49-5}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Subtract $5$ from $49$ to get $44$.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Multiply $7+\sqrt{5}$ and $7+\sqrt{5}$ to get $\left(7+\sqrt{5}\right)^{2}$.

$$\frac{\left(7+\sqrt{5}\right)^{2}}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(7+\sqrt{5}\right)^{2}$.

$$\frac{49+14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

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

$$\frac{49+14\sqrt{5}+5}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Add $49$ and $5$ to get $54$.

$$\frac{54+14\sqrt{5}}{44}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=x+7\sqrt{5}y$$

Rationalize the denominator of $\frac{7-\sqrt{5}}{7+\sqrt{5}}$ by multiplying numerator and denominator by $7-\sqrt{5}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}=x+7\sqrt{5}y$$

Consider $\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}=x+7\sqrt{5}y$$

Square $7$. Square $\sqrt{5}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{49-5}=x+7\sqrt{5}y$$

Subtract $5$ from $49$ to get $44$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{44}=x+7\sqrt{5}y$$

Multiply $7-\sqrt{5}$ and $7-\sqrt{5}$ to get $\left(7-\sqrt{5}\right)^{2}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{\left(7-\sqrt{5}\right)^{2}}{44}=x+7\sqrt{5}y$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(7-\sqrt{5}\right)^{2}$.

$$\frac{54+14\sqrt{5}}{44}-\frac{49-14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}=x+7\sqrt{5}y$$

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

$$\frac{54+14\sqrt{5}}{44}-\frac{49-14\sqrt{5}+5}{44}=x+7\sqrt{5}y$$

Add $49$ and $5$ to get $54$.

$$\frac{54+14\sqrt{5}}{44}-\frac{54-14\sqrt{5}}{44}=x+7\sqrt{5}y$$

Since $\frac{54+14\sqrt{5}}{44}$ and $\frac{54-14\sqrt{5}}{44}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{54+14\sqrt{5}-\left(54-14\sqrt{5}\right)}{44}=x+7\sqrt{5}y$$

Do the multiplications in $54+14\sqrt{5}-\left(54-14\sqrt{5}\right)$.

$$\frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}=x+7\sqrt{5}y$$

Do the calculations in $54+14\sqrt{5}-54+14\sqrt{5}$.

$$\frac{28\sqrt{5}}{44}=x+7\sqrt{5}y$$

Divide $28\sqrt{5}$ by $44$ to get $\frac{7}{11}\sqrt{5}$.

$$\frac{7}{11}\sqrt{5}=x+7\sqrt{5}y$$

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

$$x+7\sqrt{5}y=\frac{7}{11}\sqrt{5}$$

Subtract $x$ from both sides.

$$7\sqrt{5}y=\frac{7}{11}\sqrt{5}-x$$

The equation is in standard form.

$$7\sqrt{5}y=-x+\frac{7\sqrt{5}}{11}$$

Divide both sides by $7\sqrt{5}$.

$$\frac{7\sqrt{5}y}{7\sqrt{5}}=\frac{-x+\frac{7\sqrt{5}}{11}}{7\sqrt{5}}$$

Dividing by $7\sqrt{5}$ undoes the multiplication by $7\sqrt{5}$.

$$y=\frac{-x+\frac{7\sqrt{5}}{11}}{7\sqrt{5}}$$

Divide $-x+\frac{7\sqrt{5}}{11}$ by $7\sqrt{5}$.

$$y=-\frac{\sqrt{5}x}{35}+\frac{1}{11}$$