Question

1. \\(\frac{12}{18} = \frac{3x + 4}{15}\\)\
2. \\(\frac{6}{x + 16} = \frac{7}{3x + 3}\\)\
3. \\(\frac{2x + 5}{6} = \frac{7}{x - 6}\\)

Explanation:

Problem 16: $\boldsymbol{\frac{12}{18} = \frac{3x + 4}{15}}$

Step 1: Cross - multiply

Cross - multiply the fractions. If $\frac{a}{b}=\frac{c}{d}$, then $a\times d = b\times c$. So, $12\times15 = 18\times(3x + 4)$.
$12\times15=180$ and $18\times(3x + 4)=54x+72$. So the equation becomes $180 = 54x+72$.

Step 2: Subtract 72 from both sides

Subtract 72 from each side of the equation to isolate the term with $x$.
$180 - 72=54x+72 - 72$.
$108 = 54x$.

Step 3: Divide both sides by 54

Divide both sides of the equation by 54 to solve for $x$.
$\frac{108}{54}=\frac{54x}{54}$.
$x = 2$.

Step 1: Cross - multiply

Using the cross - multiplication rule for proportions $\frac{a}{b}=\frac{c}{d}\Rightarrow a\times d=b\times c$. So, $6\times(3x + 3)=7\times(x + 16)$.
Expand both sides: $6\times3x+6\times3=7\times x+7\times16$.
$18x + 18=7x+112$.

Step 2: Subtract 7x from both sides

Subtract $7x$ from each side to get the $x$ terms on one side.
$18x-7x + 18=7x-7x+112$.
$11x + 18=112$.

Step 3: Subtract 18 from both sides

Subtract 18 from both sides to isolate the term with $x$.
$11x+18 - 18=112 - 18$.
$11x = 94$. Wait, there is a mistake. Let's re - expand:

Wait, $6\times(3x + 3)=18x+18$ and $7\times(x + 16)=7x + 112$. Then $18x+18=7x + 112$. Subtract $7x$: $11x+18 = 112$. Subtract 18: $11x=94$? Wait, no, $112 - 18 = 94$? Wait, $112-18 = 94$? Wait, $112-10 = 102$, $102 - 8=94$. Then $x=\frac{94}{11}\approx8.55$? Wait, no, let's check the cross - multiplication again.

Wait, the original equation is $\frac{6}{x + 16}=\frac{7}{3x + 3}$. Cross - multiply: $6(3x + 3)=7(x + 16)$.

$18x+18 = 7x + 112$.

Subtract $7x$: $11x+18=112$.

Subtract 18: $11x=112 - 18=94$. Then $x=\frac{94}{11}\approx8.55$? Wait, that can't be right. Wait, maybe I made a mistake in arithmetic. Let's recalculate $7\times16 = 112$, $6\times3 = 18$, $6\times3x=18x$, $7\times x = 7x$. So the steps are correct. Wait, but let's check with $x=\frac{94}{11}$.

Left - hand side denominator: $x + 16=\frac{94}{11}+16=\frac{94 + 176}{11}=\frac{270}{11}$, $\frac{6}{\frac{270}{11}}=\frac{6\times11}{270}=\frac{66}{270}=\frac{11}{45}$.

Right - hand side denominator: $3x + 3=3\times\frac{94}{11}+3=\frac{282}{11}+\frac{33}{11}=\frac{315}{11}$, $\frac{7}{\frac{315}{11}}=\frac{7\times11}{315}=\frac{77}{315}=\frac{11}{45}$. Oh, it is correct. So $x=\frac{94}{11}$. Wait, but maybe I made a mistake in the problem interpretation. Wait, the original problem is $\frac{6}{x + 16}=\frac{7}{3x + 3}$. Let's check again.

Wait, $6(3x + 3)=7(x + 16)$

$18x+18 = 7x + 112$

$18x-7x=112 - 18$

$11x=94$

$x=\frac{94}{11}\approx8.55$

Step 1: Cross - multiply

Using the cross - multiplication rule $\frac{a}{b}=\frac{c}{d}\Rightarrow a\times d=b\times c$. So, $(2x + 5)\times(x - 6)=6\times7$.
Expand the left - hand side: $2x\times x-2x\times6+5\times x - 5\times6=42$.
$2x^{2}-12x + 5x-30 = 42$.
$2x^{2}-7x-30 = 42$.

Step 2: Subtract 42 from both sides

Subtract 42 from both sides to set the equation to zero.
$2x^{2}-7x-30 - 42=42 - 42$.
$2x^{2}-7x - 72=0$.

Step 3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$, the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 2$, $b=-7$, $c=-72$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-7)^{2}-4\times2\times(-72)=49 + 576=625$.
Then $x=\frac{-(-7)\pm\sqrt{625}}{2\times2}=\frac{7\pm25}{4}$.
We have two solutions:
$x_1=\frac{7 + 25}{4}=\frac{32}{4}=8$ and $x_2=\frac{7-25}{4}=\frac{-18}{4}=-\frac{9}{2}$.
We need to check for extraneous solutions (since the original equation has denominators, we need to ensure that $x-6
eq0$ and $6
eq0$ (6 is never zero)).
For $x = 8$: $x-6=8 - 6 = 2
eq0$, so it is valid.
For $x=-\frac{9}{2}$: $x - 6=-\frac{9}{2}-6=-\frac{9 + 12}{2}=-\frac{21}{2}
eq0$, so it is also valid.

Answer:

$x = 2$

Problem 18: $\boldsymbol{\frac{6}{x + 16}=\frac{7}{3x + 3}}$