Question

system of equations using substitution method
3x+4y=-3 x+2y=-1​

Answer 1

Answer:

${x,y}={-1,0}$

Step-by-step explanation:

System of Linear Equations entered :

[1]    3x + 4y = -3

  [2]    x + 2y = -1

Solve by Substitution :

// Solve equation [2] for the variable x

[2]    x = -2y - 1

// Plug this in for variable  x  in equation [1]

  [1]    3•(-2y-1) + 4y = -3

  [1]     - 2y = 0

// Solve equation [1] for the variable  y

  [1]    2y = 0

  [1]    y = 0

// By now we know this much :

   x = -2y-1

   y = 0

// Use the  y  value to solve for  x

   x = -2(0/32765)-1 = -1

Solution :

{x,y} = {-1,0/32765}

Answer 2

Answer:

$\boxed{\sf{x=-1 \quad y=0 \quad (-1,0)}}$

Step-by-step explanation:

Isolate the term of x and y from one side of the equation.

3x+4y=-3 and x+2y=-1

$\Longrightarrow: \sf{3x+4y=-3 \quad x=\dfrac{-3-4y}{3} }$

You have to substitute.

$:\Longrightarrow \sf{\dfrac{-3-4y}{3}+2y=-1}$

Solve.

$\sf{\dfrac{-3+2y}{3}=-1}$

$\sf{\dfrac{-3-4y}{3}}$

y=0

For y=0.

$\Longrightarrow: \sf{x=\dfrac{-3-4*0}{3}}$

Solve.

PEMDAS stands for:

• Parentheses
• Exponents
• Multiply
• Divide
• Add
• Subtract

-3-4*0/3

Multiply.

4*0=0

-3-0

Add/subtract the numbers from left to right.

-3-0=-3

-3/3

Divide.

-3/3=-1

x=-1

Therefore, the final answer is x=-1 and y=0.

I hope this helps. Let me know if you have any questions.