To solve for the perimeter of the triangle with sides represented by the given expressions [tex]\( \left(\frac{7}{8}x + \frac{3}{4} y\right) \)[/tex], [tex]\( (2y - 1) \)[/tex], and [tex]\( \left( y - \frac{1}{2} x + \frac{2}{3} \right) \)[/tex], we need to find the sum of these expressions.
Let's add the three expressions step-by-step:
1. The first side is given as [tex]\( \frac{7}{8} x + \frac{3}{4} y \)[/tex].
2. The second side is [tex]\( 2y - 1 \)[/tex].
3. The third side is [tex]\( y - \frac{1}{2} x + \frac{2}{3} \)[/tex].
Now, we'll add these expressions together:
[tex]\[ \left( \frac{7}{8} x + \frac{3}{4} y \right) + (2y - 1) + \left( y - \frac{1}{2} x + \frac{2}{3} \right) \][/tex]
First, let's combine the terms involving [tex]\( x \)[/tex]:
[tex]\[ \frac{7}{8} x - \frac{1}{2} x \][/tex]
To combine these, we need a common denominator, which for 8 and 2 is 8. Convert [tex]\(- \frac{1}{2} x\)[/tex] to have the same denominator:
[tex]\[ \frac{7}{8} x - \frac{4}{8} x = \frac{3}{8} x \][/tex]
Next, let's combine the terms involving [tex]\( y \)[/tex]:
[tex]\[ \frac{3}{4} y + 2y + y \][/tex]
To combine these, we convert [tex]\( 2y \)[/tex] and [tex]\( y \)[/tex] into fractions with denominator 4:
[tex]\[ \frac{3}{4} y + \frac{8}{4} y + \frac{4}{4} y = \frac{15}{4} y \][/tex]
Now, let's combine the constant terms:
[tex]\[ -1 + \frac{2}{3} \][/tex]
To add these, we need a common denominator, which for 1 and 3 is 3. Convert [tex]\(-1\)[/tex] to have the same denominator:
[tex]\[ - \frac{3}{3} + \frac{2}{3} = - \frac{1}{3} \][/tex]
Adding these combined terms together, the perimeter [tex]\( P \)[/tex] of the triangle is:
[tex]\[ P = \frac{3}{8} x + \frac{15}{4} y - \frac{1}{3} \][/tex]
Thus, the perimeter of the triangle, in simplified form, is:
[tex]\[ \boxed{\frac{3}{8} x + \frac{15}{4} y - \frac{1}{3}} \][/tex]
