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Solve the system by substitution.

minus, 5, y, minus, 10, equals, x
−5y−10=
x
x, plus, 9, y, equals, minus, 2
x+9y=
−2

7 answers

To solve the system of equations by substitution, we start with the given equations:

  1. \(-5y - 10 = x\) (we can rearrange this to express \(x\) in terms of \(y\))
  2. \(x + 9y = -2\)

From equation 1, we can express \(x\) as: \[ x = -5y - 10 \]

Now, we can substitute this expression for \(x\) into equation 2: \[ (-5y - 10) + 9y = -2 \]

Now, simplify the left side: \[ -5y - 10 + 9y = -2 \] Combine like terms: \[ (9y - 5y) - 10 = -2 \] \[ 4y - 10 = -2 \]

Next, add 10 to both sides: \[ 4y = -2 + 10 \] \[ 4y = 8 \]

Now, divide by 4: \[ y = 2 \]

Now that we have \(y\), we will substitute it back into the equation for \(x\): \[ x = -5(2) - 10 \] \[ x = -10 - 10 \] \[ x = -20 \]

So, the solution to the system of equations is: \[ (x, y) = (-20, 2) \]

To verify our solution, we can substitute \(x = -20\) and \(y = 2\) back into the original equations:

  1. For the first equation: \[ -5(2) - 10 = -20 \quad \text{(True)} \]
  2. For the second equation: \[ -20 + 9(2) = -20 + 18 = -2 \quad \text{(True)} \]

Both equations are satisfied, confirming that the solution is correct.

Final answer: \[ (x, y) = (-20, 2) \]

Solve the system by substitution.
minus, 4, x, plus, 6, y, equals, minus, 10
−4x+6y=
−10
2, y, plus, 4, equals, x
2y+4=
x

To solve the system of equations by substitution, we start with the given equations:

  1. \(-4x + 6y = -10\)
  2. \(x = 2y + 4\)

Since the second equation already expresses \(x\) in terms of \(y\), we can substitute this expression for \(x\) into the first equation.

Substituting \(x\) in the first equation: \[ -4(2y + 4) + 6y = -10 \]

Expanding the equation: \[ -8y - 16 + 6y = -10 \]

Combining like terms: \[ -8y + 6y - 16 = -10 \] \[ -2y - 16 = -10 \]

Adding 16 to both sides: \[ -2y = -10 + 16 \] \[ -2y = 6 \]

Dividing both sides by -2: \[ y = -3 \]

Now that we have \(y\), we will substitute it back into the equation for \(x\): \[ x = 2(-3) + 4 \] \[ x = -6 + 4 \] \[ x = -2 \]

So, the solution to the system of equations is: \[ (x, y) = (-2, -3) \]

Verification:

To verify our solution, we can substitute \(x = -2\) and \(y = -3\) back into the original equations:

  1. For the first equation: \[ -4(-2) + 6(-3) = 8 - 18 = -10 \quad \text{(True)} \]
  2. For the second equation: \[ -2 = 2(-3) + 4 = -6 + 4 = -2 \quad \text{(True)} \]

Both equations are satisfied, confirming that the solution is correct.

Final answer: \[ (x, y) = (-2, -3) \]

Solve the following system of equations algebraically:
y, equals, x, squared, minus, x, minus, 65
y=
x
2
−x−65
y, equals, minus, x, minus, 1
y=
−x−1
Answer
Attempt 1 out of 2

Two solutions
and and

To solve the system of equations:

  1. \( y = x^2 - x - 65 \)
  2. \( y = -x - 1 \)

we can set the two equations equal to each other since they both equal \(y\):

\[ x^2 - x - 65 = -x - 1 \]

Next, we'll add \(x + 1\) to both sides to eliminate the right-hand side:

\[ x^2 - x + x - 65 + 1 = 0 \]

This simplifies to:

\[ x^2 - 64 = 0 \]

Now, we can factor or solve this quadratic equation. This can be factored as follows:

\[ (x - 8)(x + 8) = 0 \]

Setting each factor equal to zero gives us two solutions for \(x\):

\[ x - 8 = 0 \quad \Rightarrow \quad x = 8 \] \[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \]

Now that we have the \(x\) values, we can find the corresponding \(y\) values by substituting each \(x\) back into either original equation. We'll use the second equation for simplicity:

For \(x = 8\): \[ y = -8 - 1 = -9 \]

For \(x = -8\): \[ y = -(-8) - 1 = 8 - 1 = 7 \]

Thus, the two solutions are:

  1. \( (8, -9) \)
  2. \( (-8, 7) \)

So the final answer includes both solutions: \[ (8, -9) \quad \text{and} \quad (-8, 7) \]

Solve the following system of equations algebraically:
y, equals, x, squared, minus, 7, x, minus, 7
y=
x
2
−7x−7
y, equals, minus, 3, x, minus, 2
y=
−3x−2
Answer
Attempt 1 out of 2

Two solutions
and and

To solve the system of equations:

  1. \( y = x^2 - 7x - 7 \)
  2. \( y = -3x - 2 \)

we can set the two expressions for \(y\) equal to each other since both equations are equal to \(y\):

\[ x^2 - 7x - 7 = -3x - 2 \]

Next, let's move all terms to one side of the equation:

\[ x^2 - 7x + 3x - 7 + 2 = 0 \]

Combine like terms:

\[ x^2 - 4x - 5 = 0 \]

Now we can factor this quadratic equation:

\[ (x - 5)(x + 1) = 0 \]

Setting each factor equal to zero gives us the solutions for \(x\):

  1. \( x - 5 = 0 \) → \( x = 5 \)
  2. \( x + 1 = 0 \) → \( x = -1 \)

Now, we can find the corresponding \(y\) values by substituting each \(x\) value back into either original equation. We'll use the second equation for simplicity:

For \(x = 5\): \[ y = -3(5) - 2 = -15 - 2 = -17 \]

For \(x = -1\): \[ y = -3(-1) - 2 = 3 - 2 = 1 \]

Thus, the two solutions are:

  1. \( (5, -17) \)
  2. \( (-1, 1) \)

So the final answer with both solutions is: \[ (5, -17) \quad \text{and} \quad (-1, 1) \]