Question

m= b= y= 2/3 7/8 plug that into a equation

Explanation:

Step1: Identify Variables

We have \( M = \frac{2}{3} \), \( B = \frac{7}{8} \), and \( Y \) (assuming \( Y \) is also a variable, but since only two values are given, maybe we are plugging \( M \) and \( B \) into an equation. Let's assume a simple equation like \( M + B + Y =? \) or \( M \times B \times Y =? \), but since \( Y \) is not given, maybe the equation is in terms of \( M \), \( B \), and \( Y \), and we plug \( M = \frac{2}{3} \), \( B = \frac{7}{8} \). Let's take a general equation, say \( E = M + B + Y \). Plugging \( M = \frac{2}{3} \), \( B = \frac{7}{8} \), we get \( E=\frac{2}{3}+\frac{7}{8}+Y \). First, find a common denominator for \( \frac{2}{3} \) and \( \frac{7}{8} \), which is 24.

Step2: Add Fractions

\( \frac{2}{3}=\frac{2\times8}{3\times8}=\frac{16}{24} \), \( \frac{7}{8}=\frac{7\times3}{8\times3}=\frac{21}{24} \). Then \( \frac{16}{24}+\frac{21}{24}=\frac{16 + 21}{24}=\frac{37}{24} \). So \( E=\frac{37}{24}+Y \). If \( Y \) is a variable, this is the equation with \( M \) and \( B \) plugged in. If \( Y \) was a typo and we just need to plug \( M \) and \( B \) into an equation, say \( M\times B \), then \( \frac{2}{3}\times\frac{7}{8}=\frac{14}{24}=\frac{7}{12} \). Assuming the equation is \( M\times B\times Y \) (but \( Y \) is unknown), or a simpler equation with \( M \) and \( B \). Let's assume the equation is \( M + B \) (ignoring \( Y \) as a typo). Then \( \frac{2}{3}+\frac{7}{8}=\frac{16 + 21}{24}=\frac{37}{24}=1\frac{13}{24} \).

Answer:

If the equation is \( M + B \), the result is \( \frac{37}{24} \) (or \( 1\frac{13}{24} \)). If the equation is \( M\times B \), the result is \( \frac{7}{12} \). (Since the equation is not specified, we can show both common cases. If we assume the equation is \( M + B + Y \) with \( Y \) unknown, the expression is \( \frac{37}{24}+Y \).)