The problem asks us to find the value of the variables in the given equations. We need to solve for $y$ in the equation $8y + 4 = 28$, for $m$ in the equation $m/5 = 9$, for $t$ in the equation $11 + t = 20$, for $m$ in the equation $5m + 6 - 2m = 21$, and for $f$ in the equation $5f + 8 = 50$.

AlgebraLinear EquationsSolving EquationsVariables
2025/6/16

1. Problem Description

The problem asks us to find the value of the variables in the given equations. We need to solve for yy in the equation 8y+4=288y + 4 = 28, for mm in the equation m/5=9m/5 = 9, for tt in the equation 11+t=2011 + t = 20, for mm in the equation 5m+62m=215m + 6 - 2m = 21, and for ff in the equation 5f+8=505f + 8 = 50.

2. Solution Steps

* Problem 1: 8y+4=288y + 4 = 28
Subtract 4 from both sides of the equation:
8y=2848y = 28 - 4
8y=248y = 24
Divide both sides by 8:
y=248y = \frac{24}{8}
y=3y = 3
* Problem 2: 11+t=2011 + t = 20
Subtract 11 from both sides of the equation:
t=2011t = 20 - 11
t=9t = 9
* Problem 3: m5=9\frac{m}{5} = 9
Multiply both sides of the equation by 5:
m=95m = 9 * 5
m=45m = 45
* Problem 4: 5m+62m=215m + 6 - 2m = 21
Combine like terms:
3m+6=213m + 6 = 21
Subtract 6 from both sides of the equation:
3m=2163m = 21 - 6
3m=153m = 15
Divide both sides by 3:
m=153m = \frac{15}{3}
m=5m = 5
* Problem 5: 5f+8=505f + 8 = 50
Subtract 8 from both sides of the equation:
5f=5085f = 50 - 8
5f=425f = 42
Divide both sides by 5:
f=425f = \frac{42}{5}
f=8.4f = 8.4

3. Final Answer

1. $y = 3$

2. $t = 9$

3. $m = 45$

4. $m = 5$

5. $f = \frac{42}{5}$ or $f = 8.4$

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