Question

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

Answer 1

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

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

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

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=(-3-4y)/(3) }`

You have to substitute.

`:\Longrightarrow \sf{(-3-4y)/(3)+2y=-1}`

Solve.

`\sf{(-3+2y)/(3)=-1}`

`\sf{(-3-4y)/(3)}`

y=0

For y=0.

`\Longrightarrow: \sf{x=(-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.