Linear Equations with Two Variables: Applying the formula

Note that in at least one of the equations, one of the variables has a coefficient of 1, therefore it will be easy to isolate one of the variables from one of the equations and use substitution method to easily obtain an equation with one variable.

Let's solve, then, the system of equations:

From the first equation we will isolate one of the variables (in this problem we choose to isolate y, similarly we could choose to isolate the other variable):

$x+y=8 \\ y=8-x$

We isolated the variable y by keeping it alone on the left side, this was done by moving the second term (x) to the other side, we did this while remembering that a term changes its sign when crossing sides,

Let's examine now the current system of equations:

$\begin{cases} \bm{y=\underline{8-x} } \\ x-\underline{y}=6 \end{cases}$

Now we'll substitute the entire expression that y equals from the first equation in place of y in the second equation (marked with underline in both equations above) and thus we'll get one equation with one variable:

$x-\underline{(8-x)}=6$

where:

a. We did this carefully using parentheses, because we substituted a complete expression for the variable, which in the second equation has a coefficient that isn't 1 (in this case minus 1, but for any coefficient, we'll always use parentheses when substituting). Note the substitution we performed using the underline in the last equation we got above.

b. We'll highlight the equation where the variable we substituted is isolated in order to return to it later after we find the value of x from solving the equation we got, and this is to find the value of y corresponding to that x value we'll find, therefore we highlighted this equation above.

From here - we'll continue and solve the one-variable equation we got, first we'll distribute using the distributive property:

$x-(8-x)=6 \\ x-1\cdot8-1\cdot(-x)=6\\ x-8+x=6$$$

Now we'll combine like terms and isolate x (and its coefficient) on the left side, this we'll do by moving the other terms to the right side:

$x-8+x=6 \\ 2x-8=6\\ 2x=6+8\\ 2x=14$$$

We need to ensure that x's coefficient is 1, we'll do this by dividing both sides of the equation by its coefficient, meaning we'll divide the equation by 2:

$2x=14 \hspace{8pt} \text{/:} 2 \\ \frac{\not{2}x}{\not{2}}=\frac{14}{2}\\ x=7$

where in the first stage we divided both sides of the equation by x's coefficient from the last equation, then we wrote the division result using a fraction and then we reduced the fraction on the left side and calculated (also using reduction actually) the division result on the right side.

We got the value of x that solves the system of equations above,

Now we'll return to the equation where the second variable - y is isolated and given as a function of x which we highlighted earlier:

$y=8-x$$$

and we'll substitute in it the value of x that we got earlier to find the corresponding value of y:

$y=8-\underline{x} \\ x=\underline{7}\\ \hspace{15pt}\downarrow\\ y=8-\underline{7}\\ y=1$

where in the last stage we combined like terms on the right side of the equation we got for y,

Therefore we got that the solution is:

$x=7,\hspace{8pt}y=1$

or written as an ordered pair:

$(x,y)\rightarrow (7,1)$

Therefore the correct answer is answer b.

Note:

The solution is the pair:$(x,y)=(7,1)$

meaning- only substituting both variables' values together, in both original equations (or any of the equivalent equations with two variables that we got on the way to the solution) will result in a true statement.

To solve this system of equations, we'll employ the elimination method.

The given system is:

• $-x + y = 14$

• $5x + 2y = 7$

To eliminate $y$, we can multiply the first equation by $2$ to match the coefficient of $y$ in the second equation:

$2(-x + y) = 2 \times 14$

Resulting in:

$-2x + 2y = 28$

Now, we have:

• $-2x + 2y = 28$

• $5x + 2y = 7$

Subtract the second equation from the first:

$\begin{aligned} (-2x + 2y) - (5x + 2y) &= 28 - 7 \\ -2x + 2y - 5x - 2y &= 21 \\ -7x &= 21 \end{aligned}$

Solving for $x$:

$x = \frac{21}{-7} = -3$

Next, substitute $x = -3$ back into the first equation:

$-(-3) + y = 14$

$3 + y = 14$

Solving for $y$:

$y = 14 - 3 = 11$

Therefore, the solution to the system of equations is $x = -3$ and $y = 11$.

The correct answer choice is:

$x=-3,y=11$

Note that in the current system of equations, one of the variables is isolated alone on the left side of the equation:

$\begin{cases} \underline{x}+y=8 \\ \bm{x=\underline{5-y}} \\ \end{cases}$

Therefore, we can apply the substitution method and substitute the entire expression that x equals in the second equation in place of x in the first equation (marked with an underline in both equations above) Hence we obtain one equation with one variable:

$\underline{ 5-y}+y=8$

Highlight the equation in which the variable we substituted is isolated in order to return to it later.

From here - we'll proceed to solve the single-variable equation that we obtained.

First- combine like terms on the left side of the resulting equation:

$5-y+y=8 \\ 5=8$$$

Note that y cancelled out in the current equation and we obtained a false statement, as shown below:

$5\neq8$ meaning-

We obtained a false statement regardless of the variables' values,

We can conclude from here that the system of equations has no solution, given that no matter which values we substitute for the variables - we won't obtain a true statement in both equations together.

Therefore the correct answer is answer D.

To solve this problem, we'll use the elimination method:

• Step 1: Align the coefficients of $x$ to eliminate it.
• Step 2: Subtract one equation from the other to solve for the remaining variable.
• Step 3: Substitute the known variable back into one of the original equations to find the value of the second variable.

Now, let's work through each step:

Step 1: The given system of equations is:

$6x + 4y = 18$ (Equation 1)

$-2x + 3y = 20$ (Equation 2)

To eliminate $x$, we want the coefficients of $x$ to be equal in magnitude. Multiply Equation 2 by 3 to match the $x$ coefficient in Equation 1:

$3(-2x + 3y) = 3(20)$

This results in:

$-6x + 9y = 60$ (Equation 3)

Step 2: Add Equation 1 and Equation 3 to eliminate $x$:

$(6x + 4y) + (-6x + 9y) = 18 + 60$

This simplifies to:

$13y = 78$

Solving for $y$, we divide both sides by 13:

$y = \frac{78}{13} = 6$

Step 3: Substitute $y = 6$ back into Equation 1:

$6x + 4(6) = 18$

$6x + 24 = 18$

Subtract 24 from both sides:

$6x = 18 - 24$

$6x = -6$

Divide both sides by 6:

$x = \frac{-6}{6} = -1$

Therefore, the solution to the system of equations is $\mathbf{x = -1, y = 6}$.

To solve this system of linear equations using the elimination method, follow these steps:

• Step 1: Align the equations for elimination. These equations are already simple, but you could multiply the first equation by 3 to align coefficients for $y$.
• Step 2: Multiply the first equation by 3 to facilitate elimination of $y$:
  $3(6x + y) = 3 \times 12$
  Thus, we get:
  $18x + 3y = 36$.
• Step 3: Rewrite and subtract the second equation from this new equation:
  $18x + 3y - (3y + 2x) = 36 - 20$,
  This simplifies to:
  $16x = 16$.
• Step 4: Solve for $x$:
  $x = \frac{16}{16} = 1$.
• Step 5: Substitute $x = 1$ back into the first original equation to find $y$:
  $6(1) + y = 12$,
  $6 + y = 12$,
  $y = 12 - 6$,
  $y = 6$.

Therefore, the solution to the system of equations is $x = 1, y = 6$.

To solve the system of equations using the elimination method, follow these steps:

• Step 1: Write down the given equations:
  $3x - y = 5$ (Equation 1)
  $5x + 2y = 12$ (Equation 2)
• Step 2: Align coefficients for elimination. Here, we aim to eliminate $y$ by aligning coefficients of $y$ in both equations. Multiply Equation 1 by $2$:
  $2(3x - y) = 2(5)$, which simplifies to $6x - 2y = 10$.
• Step 3: Add the modified Equation 1 to Equation 2:
  $(6x - 2y) + (5x + 2y) = 10 + 12$.
  The $2y$ terms cancel out, leaving $11x = 22$.
• Step 4: Solve for $x$:
  $11x = 22$ simplifies to $x = 2$.
• Step 5: Substitute $x = 2$ back into Equation 1 to solve for $y$:
  $3(2) - y = 5$, which simplifies to $6 - y = 5$. Thus, $y = 1$.

Therefore, the solution to the system of equations is $x = 2$ and $y = 1$.

This corresponds to choice 2 in the provided answer choices.

To solve this problem, we'll follow these steps:

• Step 1: Write the given equations.

• Step 2: Simplify and compare the equations to each other.

• Step 3: Determine the nature of the solution.

Now, let's work through each step:

Step 1: The given system of equations is:

$x - y = 8$
$2x - 2y = 16$

Step 2: Simplify and compare the two equations:

The second equation $2x - 2y = 16$ can be divided entirely by 2 to give:

$x - y = 8$

Step 3: Observe that both equations are identical, meaning they represent the same line.

Therefore, this system of equations has infinite solutions, as every point on the line satisfies the equation.

The correct answer to the original problem is: Infinite solutions

To solve this system using the elimination method, we'll follow these steps:

• Step 1: Multiply the second equation by 4 to align the $x$-coefficients:

The second equation is $-x - 2y = 24$. Multiply it by 4:

$-4x - 8y = 96$

• Step 2: Add this to the first equation:

The first equation is $4x - 8y = 16$.

Adding the scaled second equation gives:

$(4x - 8y) + (-4x - 8y) = 16 + 96$

Solving gives:

$-16y = 112$

• Step 3: Solve for $y$:

Divide both sides by $-16$:

$y = \frac{112}{-16} = -7$

• Step 4: Substitute $y = -7$ into the original second equation:

Use $-x - 2y = 24$:

$-x - 2(-7) = 24$

$-x + 14 = 24$

Subtract 14 from both sides:

$-x = 10$

Multiply by $-1$:

$x = -10$

Thus, the solution to the system is $x = -10$ and $y = -7$.

Therefore, the solution to the problem is $(x, y) = (-10, -7)$.

To solve this problem, we'll follow these steps:

• Step 1: Solve the first equation for one variable
• Step 2: Substitute into the second equation
• Step 3: Solve for the second variable
• Step 4: Use this value to find the first variable

Now, let's work through each step:
Step 1: From the equation $x - y = 8$, solve for $x$:

$x = y + 8$

Step 2: Substitute $x = y + 8$ into the second equation $3x + 2y = 24$.

$3(y + 8) + 2y = 24$

Simplify:
$3y + 24 + 2y = 24$

Combine like terms:
$5y + 24 = 24$

Step 3: Solve for $y$:

$5y = 24 - 24$

$5y = 0$

$y = 0$

Step 4: Substitute $y = 0$ back into the expression for $x$:

$x = 0 + 8$

$x = 8$

Therefore, the solution to the system of equations is $x = 8$ and $y = 0$.

This matches choice 1.

To solve this problem, we'll follow these steps:

• Step 1: Align the coefficients to eliminate one variable by manipulating the given equations.
• Step 2: Solve for one variable using the elimination method.
• Step 3: Substitute back to find the other variable.
• Step 4: Verify the solution by substituting back into the original equations.

Let's work through each step:

Step 1: Multiply the first equation by 2 and the second equation by 3 to align the coefficients of $y$.

This gives us:

$8x + 6y = -22$ (Equation 1 multiplied by 2)
$9x - 6y = -12$ (Equation 2 multiplied by 3)

Step 2: Add the two equations to eliminate $y$.

$(8x + 6y) + (9x - 6y) = -22 - 12$
$17x = -34$

Solve for $x$:

$x = \frac{-34}{17} = -2$

Step 3: Substitute $x = -2$ back into one of the original equations to solve for $y$. Using the first equation:

$4(-2) + 3y = -11$
$-8 + 3y = -11$
$3y = -11 + 8$
$3y = -3$
$y = \frac{-3}{3} = -1$

Step 4: Verify the solution by substituting $x = -2$ and $y = -1$ into the second original equation:

$3(-2) - 2(-1) = -4$
$-6 + 2 = -4$
$-4 = -4$ which holds true.

Therefore, the solution to the system of equations is $\mathbf{(x = -2, y = -1)}$.

To solve this problem, we'll begin with the system of equations:

$\begin{aligned} & 6x - 2y = 24 \quad \text{(Equation 1)} \\ & x + 5y = 4 \quad \text{(Equation 2)} \end{aligned}$

We will use the elimination method to solve for $x$ and $y$. First, let's align the equations to eliminate $y$.

We notice that if we multiply Equation 2 by 2, it becomes easier to align coefficients with Equation 1:

$2(x + 5y) = 2 \times 4$

Simplifying gives:

$2x + 10y = 8 \quad \text{(Equation 3)}$

Now, we have:

$\begin{aligned} & 6x - 2y = 24 \\ & 2x + 10y = 8 \\ \end{aligned}$

To eliminate $y$, let's add the equations after aligning coefficients. Multiply Equation 1 by 5 and Equation 3 by 1 to eliminate $y$:

$\begin{aligned} 5(6x - 2y) &= 5(24) \\ 1(2x + 10y) &= 1(8) \\ \end{aligned}$

Which gives:

$\begin{aligned} 30x - 10y &= 120 \\ 2x + 10y &= 8 \end{aligned}$

Adding these:

$(30x - 10y) + (2x + 10y) = 120 + 8$

$32x = 128$

Solving for $x$, we divide by 32:

$x = \frac{128}{32} = 4$

Substitute $x = 4$ back into Equation 2:

$4 + 5y = 4$

Subtract 4 on both sides:

$5y = 0$

Dividing by 5 gives:

$y = 0$

Thus, we have determined the solution to the system of equations:

$x = 4, y = 0$.

To solve the given system of equations using the elimination method, we proceed as follows:

• Step 1: Align the equations to eliminate one variable.
  We have: $\begin{aligned} 5y + 3x &= 15 \quad \text{(Equation 1)} \\ -2y - 4x &= -34 \quad \text{(Equation 2)} \end{aligned}$

• Step 2: Make the coefficients of $y$ equal in magnitude by manipulating the equations.
  Multiply Equation 1 by 2 and Equation 2 by 5:
  $\begin{aligned} (2) \cdot (5y + 3x) &= 2 \cdot 15 \quad \Rightarrow \quad 10y + 6x = 30 \\ (5) \cdot (-2y - 4x) &= 5 \cdot (-34) \quad \Rightarrow \quad -10y - 20x = -170 \end{aligned}$

• Step 3: Add the modified equations to eliminate $y$.
  $\begin{aligned} (10y + 6x) + (-10y - 20x) &= 30 + (-170) \\ 0y - 14x &= -140 \\ -14x &= -140 \end{aligned}$

• Step 4: Solve for $x$.
  Dividing by $-14$:
  $\begin{aligned} x &= \frac{-140}{-14} \\ x &= 10 \end{aligned}$

• Step 5: Substitute $x = 10$ back into one of the original equations to solve for $y$.
  Using Equation 1:
  $\begin{aligned} 5y + 3(10) &= 15 \\ 5y + 30 &= 15 \\ 5y &= 15 - 30 \\ 5y &= -15 \\ y &= \frac{-15}{5} \\ y &= -3 \end{aligned}$

Therefore, the solution to the system of equations is $x = 10$ and $y = -3$.