Answer:
[tex]\textsf{(a)}\quad x = \dfrac{11}{2}[/tex]
[tex]\textsf{(b)}\quad -\dfrac{5}{2}\leq y \leq 6[/tex]
Step-by-step explanation:
Part (a)
Given equation:
[tex]\dfrac{4x+5}{3}-\dfrac{3-2x}{2}=13[/tex]
To solve for x, begin by multiplying the numerator and denominator of the first fraction by 2, and the numerator and denominator of the second fraction by 3 so that both fractions share a common denominator:
[tex]\dfrac{2\cdot (4x+5)}{2\cdot 3}-\dfrac{3\cdot (3-2x)}{3\cdot2}=13\\\\\\\\\dfrac{8x+10}{6}-\dfrac{9-6x}{6}=13[/tex]
As the fractions now have the same denominator, we can subtract the numerators:
[tex]\dfrac{8x+10-(9-6x)}{6}=13\\\\\\\\\dfrac{8x+10-9+6x}{6}=13\\\\\\\\\dfrac{14x+1}{6}=13[/tex]
Multiply both sides of the equation by 6 to eliminate the fraction:
[tex]14x+1=78[/tex]
Subtract 1 from both sides:
[tex]14x=77[/tex]
Finally, divide both sides by 14:
[tex]x = \dfrac{77}{14}=\dfrac{7 \cdot 11}{7 \cdot 2}=\dfrac{11}{2}[/tex]
Therefore, the solution is:
[tex]\Large\boxed{\boxed{x=\dfrac{11}{2}}}[/tex]
[tex]\dotfill[/tex]
Part (b)
Given inequality:
[tex]2y^2 - 7y - 30 \leq 0[/tex]
To solve the inequality, begin by factoring the quadratic.
[tex]2y^2 + 5y - 12y - 30 \leq 0\\\\y(2y + 5) - 6(2y + 5)\leq 0\\\\(2y + 5)(y - 6)\leq 0[/tex]
Therefore, the roots of the quadratic are:
[tex]2y+5=0 \implies y=-\dfrac{5}{2}[/tex]
[tex]y-6=0 \implies y=6[/tex]
This means that the graph of the quadratic is a parabola that intersects the y-axis at y = -5/2 and y = 6.
As the leading coefficient is positive, the parabola opens to the right. This implies that the interval for which it is less than or equal to zero is between and including the y-intercepts.
Therefore, the solution to the inequality is:
[tex]\Large\boxed{\boxed{-\dfrac{5}{2}\leq y \leq 6}}[/tex]
