Question

At a clothing store Ted bought 4 shirts and 2 ties for a total price of $95. At the same store, Stephen bought 3 shirts and 3 ties for a total pace of $84. Each shirt was the same price and each tie was the same price.
Which system of equations can be used to find s, the cost of each shirt in dollars and t the cost of each tie in dollars.
Linda bought 1 shirt and 2 ties at the same store. Whats the total price in dollars and cents of Lindas purchase.​

Answer 1

lets take shirts as x and ties as y

4x+2y=95

3x+3y=84

common in 3 and 2 is 6

3*2 =6 and 2*3+6

3(4x+2y=95)

2(3x+3y=84)

12x+6y=285

6x+6y+168

add 12x and 6x  erase 6y's and add 285 and 168

18x=453

x=453/18

25.17

6x+6y=168

151.02+6y=168

6y=16.08

y=2.68

then

25.17+5.36

=30.53

total answer

Answer 2

Answer:

t(ties) = $8,50 and s(shirts) = $19,50

Linda's purchase total price $36,50

Step-by-step explanation:

In mathematics, a system of linear equations is a set of two or more linear equations with more than one unknown that make up a mathematical problem that consists of finding the values ​​of the unknowns that satisfy those operations.  the unknowns are usually represented with letters of the alphabet

For this example, we can represent as a system of linear equations as follows:

For Ted

$4s+2t=95$ where s are shirts, t are ties, and 95 the total price of both products in $.

For Stephen

$3s+3t=84$ where s are shirts, t are ties, and 84 the total price of both products in $.

Each shirt and each tie were buying in the same clothing store which means that it has the same price and we can and we can relate both equations with a system of equations as follows:

$\left \{ {{4s+2t=95} \atop {3s+3t=84}} \right.$

Using the reduction method, which is to clear in one of the equations with any unknown.

In this case the system has two unknowns, the selected one must be replaced by its equivalent value in the other equation.

Let's clear s in the second equation

$3s+3t=84\\3s=84-3t\\s=\frac{84-3t}{3} \\s=28-t$

Substituting the value of $s=28-t$ in the first equation

$4s+2t=95$ with $s=28-t$

$4(28-t)+2t=95\\112-4t+2t=95\\-2t=95-112\\t=\frac{-17}{-2}=8.50$

We got that the price of a tie is t = $8,50

Substituting the value of t in the first equation in order to obtain the value of s

$3s+3t=84\\3s+3(8.50)=84\\3s+25.50=84\\s=\frac{84-25.50}{3} = 19.50$

We got that the price of a shirt is s = $19,50

Checking if the result satisfies the equations

With t = 8.50 and s = 19.50

$\left \{ {{4s+2t=95} \atop {3s+3t=2}} \right.\\\left \{ {{4(19.50)+2(8.50)=95} \atop {3(19.50)+3(8.50)=84}} \right.\\\left \{ {{78+17=95} \atop {58.50+25.50=84}} \right.$

Linda bought 1 shirt and 2 ties at the same store. What's the total price in dollars and cents of Linda's purchase?

If a shirt value is s = $19,50 and a tie value is t = $8,50

Then s + 2t = ?

$19,50 + 2($8,50) = $36,50