2021-12-26

To solve: $\frac{4{t}^{2}}{5}=\frac{t}{5}+\frac{3}{10}$

Chanell Sanborn

Step 1
Given equation: $\frac{4{t}^{2}}{5}=\frac{t}{5}+\frac{3}{10}$
Multiply by 10 on both sides,
$10×\frac{4{t}^{2}}{5}=\left(\frac{t}{5}+\frac{3}{10}\right)10$
By distributive property,
$\left(\frac{t}{5}+\frac{3}{10}\right)10=10×\frac{t}{5}+10×\frac{3}{10}$
$\left(\frac{t}{5}+\frac{3}{10}\right)10=2t+3$
$10×\frac{4{t}^{2}}{5}=8{t}^{2}$
$8{t}^{2}=2t+3$
Subtract $-2t-3$ on both sides,
$8{t}^{2}-2t-3=2t+3-2t-3$
$8{t}^{2}-2t-3=0$
Split the middle term $-2t$ as $-6t$ and $+4t$
Replace $-2t$ with $-6t+4t$ in $8{t}^{2}-2t-3=0$
$8{t}^{2}-6t+4t-3=0$
Step 2
The above equation can be written as $2t×4t-2t×3+1×4t-3×1$
Apply distributive property for first two terms and last two terms,
$2t×4t-2t×3+1×4t-3×1=2t\left(4t-3\right)+1\left(4t-3\right)$
Again apply distributive property for $2t\left(4t-3\right)+1\left(4t-3\right)$
$2t\left(4t-3\right)+1\left(4t-3\right)=\left(2t+1\right)\left(4t-3\right)$
So the factors are $\left(2t+1\right)$ and $\left(4t-3\right)$
$8{t}^{2}-2t-3=\left(2t+1\right)\left(4t-3\right)$
To find the value of t,
Take $4t-3=0$
$4t-3+3=0+3$
$4t=3$
Divide by 4 on both sides,
$\frac{4t}{4}=\frac{3}{4}$

Durst37

Step 1
Find Least Common Multiplier of 5, 10: 10
Multiply by LCM=10
$\frac{4{t}^{2}}{5}×10=\frac{t}{5}×10+\frac{3}{10}×10$
Simplify
$8{t}^{2}=2t+3$
Subtract 3 from both sides
$8{t}^{2}-3=2t+3-3$
Simplify
$8t-2t-3=0$

$\sqrt{{\left(-2\right)}^{2}-4×8\left(-3\right)}=10$

Separate the solutions
${t}_{1}=\frac{-\left(-2\right)+10}{2×8}$
${t}_{2}=\frac{-\left(-2\right)-10}{2×8}$
${t}_{1}=$ Apply rule $-\left(a\right)=a$
$=\frac{2+10}{2×8}$
Add the numbers: $2+10=12$
$=\frac{12}{2×8}$
Multiply the numbers: $2×8=16$
$=\frac{12}{16}$
Cancel the common factor: 4
$=\frac{3}{4}$
${t}_{2}=$
$=\frac{2-10}{2×8}$
$2-10=-8$
$=\frac{-8}{2×8}$
$2×8=16$
$=\frac{-8}{16}$
Apply the fraction rule: $\frac{-a}{b}=-\frac{a}{b}$
$=-\frac{8}{16}$
Cancel the common factor: 8
$=-\frac{1}{2}$
The solutions to the quadratic equation are:

karton

Given: $\frac{4{t}^{2}}{5}=\frac{t}{5}+\frac{3}{10}$
Multiply both sides of the equation by 10, the least common multiple of 5,10
$2×4{t}^{2}=2t+3$
Multiply 2 and 4 to get 8.
$8{t}^{2}=2t+3$
Subtract 2t from both sides.
$8{t}^{2}-2t=3$
Subtract 3 from both sides.
$8{t}^{2}-2t-3=0$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as
$8{t}^{2}+at+bt-3$
To find a and b, set up a system to be solved.
a+b=-2
ab=8(-3)=-24
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.
1, -24
2, -12
3, -8
4, -6
Calculate the sum for each pair.
1-24=-23
2-12=-10
3-8=-5
4-6=-2
The solution is the pair that gives sum -2.
a=-6
b=4
Rewrite $8{t}^{2}-2t-3$ as $\left(8{t}^{2}-6t\right)+\left(4t-3\right)$
$\left(8{t}^{2}-6t\right)+\left(4t-3\right)$
Factor out 2t in $8{t}^{2}-6t$
2t(4t-3)+4t-3
Factor out common term 4t-3 by using distributive property.
(4t-3)(2t+1)
To find equation solutions, solve 4t-3=0 and 2t+1=0.
$t=\frac{3}{4}$
$t=-\frac{1}{2}$

Do you have a similar question?
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 81, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}