Math  /  Algebra

Question2. 25y+32x=132×522=y2+3x=1\begin{array}{l}\frac{2}{5} y \frac{+3}{2} x=1 \quad \frac{3}{2} \times \frac{5}{2} \\ \quad \frac{2}{=} y \stackrel{+3}{2} x=1\end{array}

Studdy Solution
Discuss the equation 25y+32x=1 \frac{2}{5}y + \frac{3}{2}x = 1 :
Since this is a linear equation with two variables, it represents a line in the xy xy -plane. Without additional constraints or a second equation, we cannot solve for unique values of x x and y y . However, we can express one variable in terms of the other.
For example, solving for y y in terms of x x :
25y=132x \frac{2}{5}y = 1 - \frac{3}{2}x
Multiply both sides by 52 \frac{5}{2} to isolate y y :
y=52(132x) y = \frac{5}{2} \left( 1 - \frac{3}{2}x \right)
y=52154x y = \frac{5}{2} - \frac{15}{4}x
This equation expresses y y in terms of x x .
The multiplication result is 154 \frac{15}{4} , and the equation 25y+32x=1 \frac{2}{5}y + \frac{3}{2}x = 1 can be expressed as y=52154x y = \frac{5}{2} - \frac{15}{4}x .

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