MO MAP Algebra I Algebra. Practice it free.

The greatest common factor of $10x^{2}$ and $15x$ is $5x\textsf{.}$ Factoring gives $5x(2x - 3)$.

$16x^{2} - 49 = (4x)^{2} - 7^{2} = (4x - 7)(4x + 7)$.

Write $36m$ with denominator $t$: $\dfrac{18m}{t} + \dfrac{36mt}{t} = \dfrac{18m + 36mt}{t} = \dfrac{18m(1 + 2t)}{t}$.

Factor out $2$ first: $2x^2 + 4x - 6 = 2(x^2 + 2x - 3)$. Then find two numbers with product $-3$ and sum $2$: $3$ and $-1$. So $2x^2 + 4x - 6 = 2(x+3)(x-1)$.

Half of $10$ is $5$ and $5^{2} = 25$, so $x^{2} + 10x + 25 = (x + 5)^{2}$ and $a = 5$.

Two numbers that multiply to $12$ and add to $7$ are $3$ and $4$. So $x^{2} + 7x + 12 = (x + 3)(x + 4)$.

Substitute $y = 8$. From $y \ge 2x + 1$, $8 \ge 2x + 1$, so $x \le \dfrac{7}{2}$. From $y \le x + 9$, $8 \le x + 9$, so $x \ge -1$. The value $x = 2$ satisfies both inequalities.

Set the expressions equal: $4x-3=2x^{2}-x-3$, so $0=2x^{2}-5x$. Then $x(2x-5)=0$, and $x=0$ or $x=\dfrac{5}{2}$. For $x=\dfrac{5}{2}$, $y=4\left(\dfrac{5}{2}\right)-3=7$, so one intersection point is $\left(\dfrac{5}{2},7\right)\textsf{.}$