Understanding how to determine the value of yyy in various mathematical contexts is fundamental in mathematics and its applications. Whether you are dealing with simple linear equations, complex algebraic expressions, or sophisticated functions, finding the value of yyy is a crucial skill. This comprehensive guide will walk you through different scenarios and methods for determining the value of yyy, providing you with a thorough understanding and practical examples.

**Linear Equations**

**Basic Linear Equations**

A **linear equation** is an equation that makes a straight line when it is graphed. The standard form of a linear equation in two variables is y=mx+by = mx + by=mx+b, where mmm is the slope and bbb is the y-intercept.

**Example:** Given the equation y=2x+3y = 2x + 3y=2x+3, to find the value of yyy when x=4x = 4x=4: y=2(4)+3y = 2(4) + 3y=2(4)+3 y=8+3y = 8 + 3y=8+3 y=11y = 11y=11

**Solving for y in Linear Equations**

In equations where you need to isolate yyy, the process involves basic algebraic manipulation. Consider the equation 3x+4y=123x + 4y = 123x+4y=12.

**Steps to solve for yyy:**

- Isolate yyy by subtracting 3x3x3x from both sides: 4y=12−3x4y = 12 – 3x4y=12−3x
- Divide both sides by 4: y=12−3x4y = \frac{12 – 3x}{4}y=412−3x y=3−3x4y = 3 – \frac{3x}{4}y=3−43x

**Quadratic Equations**

**Standard Form of Quadratic Equations**

A **quadratic equation** is represented as y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c. The graph of a quadratic equation is a parabola.

**Example:** Given y=2×2+3x+1y = 2x^2 + 3x + 1y=2×2+3x+1, to find the value of yyy when x=2x = 2x=2: y=2(2)2+3(2)+1y = 2(2)^2 + 3(2) + 1y=2(2)2+3(2)+1 y=2(4)+6+1y = 2(4) + 6 + 1y=2(4)+6+1 y=8+6+1y = 8 + 6 + 1y=8+6+1 y=15y = 15y=15

**Solving Quadratic Equations**

To solve a quadratic equation for yyy, you can use the quadratic formula y=−b±b2−4ac2ay = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}y=2a−b±b2−4ac.

**Example:** For the equation x2−4x−5=0x^2 – 4x – 5 = 0x2−4x−5=0:

- Identify a=1a = 1a=1, b=−4b = -4b=−4, and c=−5c = -5c=−5.
- Apply the quadratic formula: x=−(−4)±(−4)2−4(1)(−5)2(1)x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(-5)}}{2(1)}x=2(1)−(−4)±(−4)2−4(1)(−5) x=4±16+202x = \frac{4 \pm \sqrt{16 + 20}}{2}x=24±16+20 x=4±362x = \frac{4 \pm \sqrt{36}}{2}x=24±36 x=4±62x = \frac{4 \pm 6}{2}x=24±6 x=5orx=−1x = 5 \quad \text{or} \quad x = -1x=5orx=−1

When x=5x = 5x=5: y=2(5)2+3(5)+1y = 2(5)^2 + 3(5) + 1y=2(5)2+3(5)+1 y=2(25)+15+1y = 2(25) + 15 + 1y=2(25)+15+1 y=50+15+1y = 50 + 15 + 1y=50+15+1 y=66y = 66y=66

When x=−1x = -1x=−1: y=2(−1)2+3(−1)+1y = 2(-1)^2 + 3(-1) + 1y=2(−1)2+3(−1)+1 y=2(1)−3+1y = 2(1) – 3 + 1y=2(1)−3+1 y=2−3+1y = 2 – 3 + 1y=2−3+1 y=0y = 0y=0

**Systems of Equations**

**Solving Linear Systems**

To find the value of yyy in a system of equations, you can use substitution or elimination methods.

**Example:** Solve the system: x+y=5x + y = 5x+y=5 2x−y=12x – y = 12x−y=1

Using substitution:

- Solve the first equation for yyy: y=5−xy = 5 – xy=5−x
- Substitute yyy in the second equation: 2x−(5−x)=12x – (5 – x) = 12x−(5−x)=1 2x−5+x=12x – 5 + x = 12x−5+x=1 3x−5=13x – 5 = 13x−5=1 3x=63x = 63x=6 x=2x = 2x=2
- Substitute xxx back into y=5−xy = 5 – xy=5−x: y=5−2y = 5 – 2y=5−2 y=3y = 3y=3

**Exponential and Logarithmic Functions**

**Exponential Functions**

Exponential functions take the form y=abxy = ab^xy=abx, where aaa is a constant and bbb is the base of the exponent.

**Example:** Given y=3(2)xy = 3(2)^xy=3(2)x, to find the value of yyy when x=3x = 3x=3: y=3(2)3y = 3(2)^3y=3(2)3 y=3(8)y = 3(8)y=3(8) y=24y = 24y=24

**Logarithmic Functions**

Logarithmic functions are the inverse of exponential functions, typically written as y=logb(x)y = \log_b(x)y=logb(x).

**Example:** Given y=log2(8)y = \log_2(8)y=log2(8), to find the value of yyy: 2y=82^y = 82y=8 2y=232^y = 2^32y=23 y=3y = 3y=3

**Trigonometric Functions**

**Sine, Cosine, and Tangent**

Trigonometric functions relate angles of a triangle to the lengths of its sides. The basic trigonometric functions include sine (sin\sinsin), cosine (cos\coscos), and tangent (tan\tantan).

**Example:** Given y=sin(x)y = \sin(x)y=sin(x), to find the value of yyy when x=π2x = \frac{\pi}{2}x=2π: y=sin(π2)y = \sin\left(\frac{\pi}{2}\right)y=sin(2π) y=1y = 1y=1

**Polynomial Functions**

**Higher-Degree Polynomials**

Polynomial functions can have degrees higher than two. The general form is y=anxn+an−1xn−1+…+a1x+a0y = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0y=anxn+an−1xn−1+…+a1x+a0.

**Example:** Given y=4×3−3×2+2x−1y = 4x^3 – 3x^2 + 2x – 1y=4×3−3×2+2x−1, to find the value of yyy when x=2x = 2x=2: y=4(2)3−3(2)2+2(2)−1y = 4(2)^3 – 3(2)^2 + 2(2) – 1y=4(2)3−3(2)2+2(2)−1 y=4(8)−3(4)+4−1y = 4(8) – 3(4) + 4 – 1y=4(8)−3(4)+4−1 y=32−12+4−1y = 32 – 12 + 4 – 1y=32−12+4−1 y=23y = 23y=23

**Conclusion**

In conclusion, finding the value of yyy is a fundamental skill in mathematics that applies to various types of equations and functions. By understanding the different methods and contexts in which yyy can be determined, we can solve a wide range of mathematical problems with precision and confidence. Whether dealing with linear equations, quadratic functions, systems of equations, exponential and logarithmic functions, trigonometric functions, or higher-degree polynomials, mastering these techniques is essential for mathematical proficiency.