Question

1.5 notetaking with vocabulary (continued)
extra practice
in exercises 1 - 6, solve the literal equation for y.

1. $y - 2x = 15$ 2. $4x + y = 2$ 3. $8 = 8 + 3y$
2. $y + x = 11$ 5. $3x - y = - 4$ 6. $3x + 1 = 7 - 4y$

in exercises 7 - 12, solve the literal equation for x.

1. $y = 10x - 4x$ 8. $q = 3x + 9xz$ 9. $r = 4 + 7x - sx$
2. $y + 4x = 10x - 6$ 11. $4g + r = 2r - 2x$ 12. $3z + 8 = 12 + 3x - z$

in exercises 13 - 16, solve the formula for the indicated variable.
· 13. area of a triangle: $a=\frac{1}{2}bh$; solve for b.
· 14. volume of a cone: $v=\frac{1}{3}pi r^{2}h$; solve for h.

Explanation:

Step1: Solve \(y - 2x=15\) for \(y\)

Add \(2x\) to both sides of the equation.
\(y-2x + 2x=15 + 2x\)

Step2: Simplify

\(y=15 + 2x\)

Step3: Solve \(4x + y=2\) for \(y\)

Subtract \(4x\) from both sides of the equation.
\(4x + y-4x=2-4x\)

Step4: Simplify

\(y=2-4x\)

Step5: Solve \(y + x=11\) for \(y\)

Subtract \(x\) from both sides of the equation.
\(y+x - x=11 - x\)

Step6: Simplify

\(y=11 - x\)

Step7: Solve \(3x - y=-4\) for \(y\)

First, add \(y\) to both sides: \(3x - y+y=-4 + y\), which gives \(3x=-4 + y\).
Then, add \(4\) to both sides: \(3x + 4=-4 + y+4\)

Step8: Simplify

\(y=3x + 4\)

Step9: Solve \(3x + 1=7-4y\) for \(y\)

First, subtract \(7\) from both sides: \(3x + 1-7=7-4y-7\), which gives \(3x - 6=-4y\).
Then, divide both sides by \(-4\): \(\frac{3x - 6}{-4}=\frac{-4y}{-4}\)

Step10: Simplify

\(y=\frac{-3x + 6}{4}=\frac{-3x}{4}+\frac{3}{2}\)

Step11: Solve \(y = 10x-4x\) for \(y\)

Combine like - terms.
\(y=(10 - 4)x\)

Step12: Simplify

\(y = 6x\)

Step13: Solve \(y+4x=10x - 6\) for \(y\)

Subtract \(4x\) from both sides: \(y+4x-4x=10x - 6-4x\)

Step14: Simplify

\(y=6x - 6\)

Step15: Solve \(4g + r=2r-2x\) for \(r\)

First, subtract \(r\) from both sides: \(4g + r-r=2r-2x-r\), which gives \(4g=r - 2x\).
Then, add \(2x\) to both sides: \(4g + 2x=r-2x + 2x\)

Step16: Simplify

\(r=4g + 2x\)

Step17: Solve \(3z + 8=12 + 3x-z\) for \(z\)

First, add \(z\) to both sides: \(3z+z + 8=12 + 3x-z+z\), which gives \(4z + 8=12 + 3x\).
Then, subtract \(8\) from both sides: \(4z+8 - 8=12 + 3x-8\), so \(4z=4 + 3x\).
Finally, divide both sides by \(4\): \(z=\frac{4 + 3x}{4}=1+\frac{3x}{4}\)

Step18: Solve \(A=\frac{1}{2}bh\) for \(b\)

Multiply both sides by \(2\): \(2A=bh\).
Then, divide both sides by \(h\) (\(h
eq0\)): \(b=\frac{2A}{h}\)

Step19: Solve \(V=\frac{1}{3}\pi r^{2}h\) for \(h\)

Multiply both sides by \(3\): \(3V=\pi r^{2}h\).
Then, divide both sides by \(\pi r^{2}\) (\(\pi r^{2}
eq0\)): \(h=\frac{3V}{\pi r^{2}}\)

Answer:

1. \(y = 15+2x\)
2. \(y=2 - 4x\)
3. \(y=\frac{-3x + 6}{4}\)
4. \(y=11 - x\)
5. \(y=3x + 4\)
6. \(y=\frac{-3x + 6}{4}\)
7. \(y = 6x\)
8. (No variable to solve for as the equation \(q = 3x+9xz\) is not in the correct form for the given instructions. If solving for \(x\) (assuming a mis - label): \(x=\frac{q}{3 + 9z}\) (\(z

eq-\frac{1}{3}\)))

1. (No variable to solve for as the equation \(r = 4+7x-sx\) is not in the correct form for the given instructions. If solving for \(x\): \(x=\frac{r - 4}{7 - s}\) (\(s

eq7\)))

1. \(y=6x - 6\)
2. \(r=4g + 2x\)
3. \(z=1+\frac{3x}{4}\)
4. \(b=\frac{2A}{h}(h

eq0)\)

1. \(h=\frac{3V}{\pi r^{2}}(\pi r^{2}

eq0)\)