hadejada7x

2021-12-26

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

Chanell Sanborn

Beginner2021-12-27Added 41 answers

Step 1

Given equation:$\frac{4{t}^{2}}{5}=\frac{t}{5}+\frac{3}{10}$

Multiply by 10 on both sides,

$10\times \frac{4{t}^{2}}{5}=(\frac{t}{5}+\frac{3}{10})10$

By distributive property,

$(\frac{t}{5}+\frac{3}{10})10=10\times \frac{t}{5}+10\times \frac{3}{10}$

$(\frac{t}{5}+\frac{3}{10})10=2t+3$

$10\times \frac{4{t}^{2}}{5}=8{t}^{2}$

$8{t}^{2}=2t+3$

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$2t\times 4t-2t\times 3+1\times 4t-3\times 1$

Apply distributive property for first two terms and last two terms,

$2t\times 4t-2t\times 3+1\times 4t-3\times 1=2t(4t-3)+1(4t-3)$

Again apply distributive property for$2t(4t-3)+1(4t-3)$

$2t(4t-3)+1(4t-3)=(2t+1)(4t-3)$

So the factors are$(2t+1)$ and $(4t-3)$

$8{t}^{2}-2t-3=(2t+1)(4t-3)$

To find the value of t,

Take$4t-3=0$

Add 3 on both sides,

$4t-3+3=0+3$

Add the like terms,

$4t=3$

Divide by 4 on both sides,

$\frac{4t}{4}=\frac{3}{4}$

$$

Given equation:

Multiply by 10 on both sides,

By distributive property,

Subtract

Split the middle term

Replace

Step 2

The above equation can be written as

Apply distributive property for first two terms and last two terms,

Again apply distributive property for

So the factors are

To find the value of t,

Take

Add 3 on both sides,

Add the like terms,

Divide by 4 on both sides,

0

Durst37

Beginner2021-12-28Added 37 answers

Step 1

Find Least Common Multiplier of 5, 10: 10

Multiply by LCM=10

$\frac{4{t}^{2}}{5}\times 10=\frac{t}{5}\times 10+\frac{3}{10}\times 10$

Simplify

$8{t}^{2}=2t+3$

Subtract 3 from both sides

$8{t}^{2}-3=2t+3-3$

Simplify

$8t-2t-3=0$

Solve with the quadratic formula

$t}_{1,\text{}2}=\frac{-(-2)\pm \sqrt{{(-2)}^{2}-4\times 8(-3)}}{2\times 8$

$\sqrt{{(-2)}^{2}-4\times 8(-3)}=10$

$t}_{1,\text{}2}=\frac{-(-2)\pm 10}{2\times 8$

Separate the solutions

$t}_{1}=\frac{-(-2)+10}{2\times 8$

$t}_{2}=\frac{-(-2)-10}{2\times 8$

${t}_{1}=$ Apply rule $-\left(a\right)=a$

$=\frac{2+10}{2\times 8}$

Add the numbers:$2+10=12$

$=\frac{12}{2\times 8}$

Multiply the numbers:$2\times 8=16$

$=\frac{12}{16}$

Cancel the common factor: 4

$=\frac{3}{4}$

${t}_{2}=$

$=\frac{2-10}{2\times 8}$

$2-10=-8$

$=\frac{-8}{2\times 8}$

$2\times 8=16$

$=\frac{-8}{16}$

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:

$t=\frac{3}{4},\text{}t=-\frac{1}{2}$

Find Least Common Multiplier of 5, 10: 10

Multiply by LCM=10

Simplify

Subtract 3 from both sides

Simplify

Solve with the quadratic formula

Separate the solutions

Add the numbers:

Multiply the numbers:

Cancel the common factor: 4

Apply the fraction rule:

Cancel the common factor: 8

The solutions to the quadratic equation are:

karton

Expert2022-01-09Added 613 answers

Given:

Multiply both sides of the equation by 10, the least common multiple of 5,10

Multiply 2 and 4 to get 8.

Subtract 2t from both sides.

Subtract 3 from both sides.

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as

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

Factor out 2t in

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.

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