Answer:

the substitution method, follow these steps:

Step 1: Solve one equation for one variable in terms of the other variable.

We'll solve the second equation for y in terms of x:

4x - y =4x - y = 5

To solve for y, we'll isolate the variable by moving the terms without y to the other side of the equation:

-y = -4x + 5

Next, we'll multiply every term by -1 to remove the negative sign in front of y:

y = 4x - 5

So, we have solved the second equation for y in terms of x. Now, we can substitute this expression for y in the first equation:

3x + 2(4x - 5) = 14

Simplifying, we get:

3x + 8x - 10 = 14

Combining like terms:

11x - 10 = 14

Next, we'll isolate the variable by moving the constant term to the other side of the equation:

11x = 14 + 10

11x = 24

Finally, we'll solve for x by dividing both sides of the equation by 11:

x = 24/11

Therefore, the solution to the system of equations is x = 24/11.

Verified answer

Answer:

>>> [tex] \sf{x = \dfrac{24}{11} }[/tex]

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>>> [tex] \sf{y = \dfrac{41}{11} }[/tex]

Step-by-step explanation:

Given Equations ,

• 3x + 2y = 14

• 4x - y = 5

We have to find ,

• Value of x and y using substitution method

Solution :

Let us say ,

>>> 3x + 2y = 14 ----------- ( Equation 1 )

>>> 4x - y = 5 ----------- ( Equation 2 )

Now using Equation 2 :

[tex] \longmapsto \: \: \: \sf{4x - y = 5}[/tex]

Adding y to both sides :

[tex]\longmapsto \: \: \: \sf{4x - \cancel{ y} + \cancel{y} = 5 + y}[/tex]

[tex]\longmapsto \: \: \: \sf{4x = 5 + y}[/tex]

Dividing both sides with 4 :

[tex]\longmapsto \: \: \: \sf{ \dfrac{ \cancel{4}x}{ \cancel{4}} = \dfrac{5 + y}{4} }[/tex]

We get ,

[tex]\longmapsto \: \: \: \sf{ x = \dfrac{5 + y}{4} } - - (Equation 3)[/tex]

Now substituting this value of x in Equation 1 :

[tex] \dashrightarrow \: \: \: \sf{3x + 2y = 14}[/tex]

[tex] \dashrightarrow \: \: \: \sf{3 \left( \bold{\dfrac{5 + y}{4}} \right) + 2y = 14}[/tex]

Simplifying :

[tex]\dashrightarrow \: \: \: \sf{ {\dfrac{15 + 3y}{4}} + 2y = 14}[/tex]

Solving LHS by Taking LCM :

[tex]\dashrightarrow \: \: \: \sf{ {\dfrac{15 + 3y + 8y}{4}} = 14}[/tex]

Multiplying both sides with 4 :

[tex]\dashrightarrow \: \: \: \sf{ {\dfrac{15 + 3y + 8y}{ \cancel{4}}} \times \cancel{4}= 14 \times 4}[/tex]

We get ,

[tex] \dashrightarrow \: \: \: \sf{15 + 11y = 56}[/tex]

Subtracting 15 to both sides :

[tex]\dashrightarrow \: \: \: \sf{ \cancel{15} + 11y - \cancel{ 15 }= 56 - 15}[/tex]

[tex]\dashrightarrow \: \: \: \sf{ 11y = 41}[/tex]

Dividing both sides with 11 :

[tex]\dashrightarrow \: \: \: \sf{ \dfrac{ \cancel{11}y }{ \cancel{11}}= \dfrac{41}{11}}[/tex]

We get ,

[tex]\dashrightarrow \: \: \: \underline{\boxed{\sf{ \bold{y = \dfrac{41}{11} }}}} \: \: \: \bigstar[/tex]

• Therefore, value of y is "41/11"

For finding value of x we are substituting value of y in Equation 3 :

[tex] \dashrightarrow \: \: \: \sf{\sf{ x = \dfrac{5 + \bold{ \frac{41}{11}} }{4} }}[/tex]

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[tex]\dashrightarrow \: \: \: \sf{\sf{ x = \dfrac{ \frac{55 + 41}{11} }{4} }}[/tex]

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[tex]\dashrightarrow \: \: \: \sf{\sf{ x = \dfrac{ \frac{ 96}{11} }{4} }}[/tex]

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[tex]\dashrightarrow \: \: \: \sf{\sf{ x = \dfrac{ \cancel{96}}{11 \times \cancel{4} }}}[/tex]

We get ,

[tex]\dashrightarrow \: \: \: \underline{\sf{\sf{ \boxed{ \bold{ x = \dfrac{24}{11} }}}}} \: \: \: \bigstar[/tex]

• Therefore, value of x is "24/11"