Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the following equation for 
y.

{:[2y+5=sqrt(15-2y)],[y=]:}

Solve the following equation for y y .\newline2y+5=152yy= \begin{array}{l} 2 y+5=\sqrt{15-2 y} \\ y= \\ \end{array}

Full solution

Q. Solve the following equation for y y .\newline2y+5=152yy= \begin{array}{l} 2 y+5=\sqrt{15-2 y} \\ y= \\ \end{array}
  1. Write equation: Write down the given equation.\newlineWe have the equation 2y+5=152y2y + 5 = \sqrt{15 - 2y}.
  2. Isolate square root term: Isolate the square root term.\newlineTo solve for yy, we want to isolate the square root term on one side of the equation. We can do this by subtracting 55 from both sides of the equation.\newline2y+55=152y52y + 5 - 5 = \sqrt{15 - 2y} - 5\newlineThis simplifies to:\newline2y=152y52y = \sqrt{15 - 2y} - 5
  3. Move y terms to one side: Move all terms involving y to one side.\newlineTo make the equation easier to solve, we want all the y terms on one side. We can do this by adding 2y2y to both sides of the equation.\newline2y2y=152y5+2y2y - 2y = \sqrt{15 - 2y} - 5 + 2y\newlineThis simplifies to:\newline0=152y5+2y0 = \sqrt{15 - 2y} - 5 + 2y
  4. Rearrange equation: Rearrange the equation.\newlineWe can rearrange the equation to have the square root term on one side and the other terms on the other side.\newline152y=52y\sqrt{15 - 2y} = 5 - 2y
  5. Square both sides: Square both sides of the equation to eliminate the square root.\newline(152y)2=(52y)2(\sqrt{15 - 2y})^2 = (5 - 2y)^2\newlineThis simplifies to:\newline152y=(52y)(52y)15 - 2y = (5 - 2y)(5 - 2y)
  6. Expand right side: Expand the right side of the equation.\newlineWe need to expand the right side using the distributive property (FOIL method).\newline152y=2510y+4y215 - 2y = 25 - 10y + 4y^2
  7. Rearrange to quadratic equation: Rearrange the equation to form a quadratic equation.\newlineWe want to bring all terms to one side to form a quadratic equation.\newline4y2+10y25=04y^2 + 10y - 25 = 0
  8. Solve quadratic equation: Solve the quadratic equation.\newlineWe can solve the quadratic equation by factoring, completing the square, or using the quadratic formula. The quadratic formula is:\newliney=b±b24ac2ay = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\newlineFor our equation, a=4a = 4, b=10b = 10, and c=25c = -25.\newliney=10±10244(25)24y = \frac{{-10 \pm \sqrt{{10^2 - 4 \cdot 4 \cdot (-25)}}}}{{2 \cdot 4}}\newliney=10±100+4008y = \frac{{-10 \pm \sqrt{{100 + 400}}}}{8}\newliney=10±5008y = \frac{{-10 \pm \sqrt{{500}}}}{8}
  9. Simplify square root and solve: Simplify the square root and solve for yy.500\sqrt{500} can be simplified to 10510\sqrt{5}.y=10±1058y = \frac{{-10 \pm 10\sqrt{5}}}{{8}}We have two possible solutions for yy:y=10+1058y = \frac{{-10 + 10\sqrt{5}}}{{8}} or y=101058y = \frac{{-10 - 10\sqrt{5}}}{{8}}
  10. Check for extraneous solutions: Check for extraneous solutions.\newlineWe need to check if both solutions satisfy the original equation because squaring both sides of an equation can introduce extraneous solutions.\newlineFor y=10+1058y = \frac{-10 + 10\sqrt{5}}{8}:\newline2y+5=152y2y + 5 = \sqrt{15 - 2y}\newline2(10+1058)+5=152(10+1058)2\left(\frac{-10 + 10\sqrt{5}}{8}\right) + 5 = \sqrt{15 - 2\left(\frac{-10 + 10\sqrt{5}}{8}\right)}\newlineAfter simplifying, we check if both sides are equal.\newlineFor y=101058y = \frac{-10 - 10\sqrt{5}}{8}:\newline2y+5=152y2y + 5 = \sqrt{15 - 2y}\newline2(101058)+5=152(101058)2\left(\frac{-10 - 10\sqrt{5}}{8}\right) + 5 = \sqrt{15 - 2\left(\frac{-10 - 10\sqrt{5}}{8}\right)}\newlineAfter simplifying, we check if both sides are equal.\newlineIf one of the solutions does not satisfy the original equation, it is an extraneous solution and should be discarded.

More problems from Domain and range of square root functions: equations