Solve for y 13y^2+10y-3=5y^2
The problem provided is a quadratic equation that requires solving for the variable y. Specifically, the equation presented is of the form 'ax^2 + bx + c = dx^2', where 'a', 'b', 'c', and 'd' are coefficients, and 'x' is the variable to be solved. You are expected to manipulate the equation to isolate y and find its value(s) so that the equation holds true. This involves combining like terms, transferring all terms with the variable 'y' to one side of the equation, and then using algebraic techniques to solve for 'y'. The solution may involve factoring, completing the square, or using the quadratic formula.
$13 y^{2} + 10 y - 3 = 5 y^{2}$
Answer
Solution:
Step 1: Isolate the variable terms
Step 1.1: Subtract $5y^2$ from both sides
$13y^2 + 10y - 3 - 5y^2 = 0$
Step 1.2: Combine like terms
$8y^2 + 10y - 3 = 0$
Step 2: Factor the quadratic expression
Step 2.1: Find two numbers that multiply to $8 \cdot -3 = -24$ and add to $10$
Step 2.1.1: Break down the middle term
$8y^2 + 12y - 2y - 3 = 0$
Step 2.2: Group and factor out the common factors
Step 2.2.1: Group terms and factor
$(8y^2 - 2y) + (12y - 3) = 0$
Step 2.2.2: Factor by grouping
$2y(4y - 1) + 3(4y - 1) = 0$
Step 2.3: Factor out the common binomial
$(4y - 1)(2y + 3) = 0$
Step 3: Apply the zero product property
If a product of factors equals zero, at least one of the factors must be zero.
Step 4: Solve the first factor
Step 4.1: Set the first factor equal to zero
$4y - 1 = 0$
Step 4.2: Solve for $y$
$4y = 1$ $y = \frac{1}{4}$
Step 5: Solve the second factor
Step 5.1: Set the second factor equal to zero
$2y + 3 = 0$
Step 5.2: Solve for $y$
$2y = -3$ $y = -\frac{3}{2}$
Step 6: Combine the solutions
The solutions are $y = \frac{1}{4}$ and $y = -\frac{3}{2}$.
Step 7: Present the solution in various forms
Exact Form: $y = \frac{1}{4}, -\frac{3}{2}$ Decimal Form: $y = 0.25, -1.5$ Mixed Number Form: $y = \frac{1}{4}, -1\frac{1}{2}$
Knowledge Notes:
To solve the quadratic equation $13y^2 + 10y - 3 = 5y^2$, we first bring all terms to one side to set the equation to zero, which is a standard form for solving quadratics. We then factor the quadratic, looking for two numbers that multiply to the product of the leading coefficient and the constant term ($ac$ in $ax^2 + bx + c$), and add up to the middle coefficient ($b$). After factoring by grouping, we use the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This property allows us to set each factor equal to zero and solve for the variable $y$. The solutions are presented in exact form, decimal form, and mixed number form to accommodate different preferences for representing numbers.
