\begin{aligned} 4y^2+9&=6x+3\\ 4y&=2x+1 \end{aligned}

Question

\begin{aligned}   4y^2+9&=6x+3\\ 4y&=2x+1 \end{aligned}

Bytelearn · Accepted Answer

Simplify first equation: To solve the system of equations, we first look at the two equations given: \begin{aligned}  4y^2 + 9 &= 6x + 3, \ 4y &= 2x + 1. \end{aligned} We can simplify the first equation by subtracting 3 from both sides to isolate the term with y. \begin{aligned}  4y^2 + 9 - 3 &= 6x + 3 - 3, \ 4y^2 + 6 &= 6x. \end{aligned}Divide by 6: Now, we can divide the entire equation by 6 to simplify it further: \begin{aligned}  \frac{4y^2 + 6}{6} &= \frac{6x}{6}, \ \frac{2}{3}y^2 + 1 &= x. \end{aligned}Substitute x into y: Next, we can substitute the expression for x from the second equation into the first equation: \begin{aligned}  4y &= 2x + 1, \ 4y &= 2(\frac{2}{3}y^2 + 1) + 1. \end{aligned}Distribute on right side: Now, we distribute the 2 on the right side of the equation: \begin{aligned}  4y &= 2 \cdot \frac{2}{3}y^2 + 2 \cdot 1 + 1, \ 4y &= \frac{4}{3}y^2 + 2 + 1, \ 4y &= \frac{4}{3}y^2 + 3. \end{aligned}Get all terms with y: To solve for y, we need to get all terms involving y on one side and the constants on the other side. We can subtract 4y from both sides: \begin{aligned}  4y - 4y &= \frac{4}{3}y^2 + 3 - 4y, \ 0 &= \frac{4}{3}y^2 - 4y + 3. \end{aligned}Multiply by 3: We can multiply the entire equation by 3 to eliminate the fraction: \begin{aligned}  3 \cdot 0 &= 3 \cdot (\frac{4}{3}y^2 - 4y + 3), \ 0 &= 4y^2 - 12y + 9. \end{aligned}Use quadratic formula for y: Now we have a quadratic equation in standard form. We can use the quadratic formula to solve for y: \begin{aligned}  y &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \ y &= \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot 9}}{2 \cdot 4}, \ y &= \frac{12 \pm \sqrt{144 - 144}}{8}, \ y &= \frac{12 \pm \sqrt{0}}{8}, \ y &= \frac{12 \pm 0}{8}, \ y &= \frac{12}{8}, \ y &= \frac{3}{2}. \end{aligned}Substitute y into x: Now that we have the value of y, we can substitute it back into the second equation to find x: \begin{aligned}  4y &= 2x + 1, \ 4 \cdot \frac{3}{2} &= 2x + 1, \ 6 &= 2x + 1. \end{aligned}Isolate x term: Subtract 1 from both sides to isolate the term with x: \begin{aligned}  6 - 1 &= 2x + 1 - 1, \ 5 &= 2x. \end{aligned}Divide by 2: Finally, divide both sides by 2 to solve for x: \begin{aligned}  \frac{5}{2} &= x. \end{aligned}

$\begin{aligned} 4y^2+9&=6x+3\ 4y&=2x+1 \end{aligned}$

Full solution

More problems from Csc, sec, and cot of special angles

Related topics

4y2+9amp;=6x+3 4yamp;=2x+1\begin{aligned} 4y^2+9&amp;=6x+3\ 4y&amp;=2x+1 \end{aligned}4y2+9​amp;=6x+3 4y​amp;=2x+1​

$\begin{aligned} 4y^2+9&=6x+3\ 4y&=2x+1 \end{aligned}$