Linear Equations Two Variables Practice Problems Solutions

To solve the system of equations:

$5x+y=21$

$x - y = 3$

Let's solve the second equation for $x$:

$x = y + 3$

Substitute $x = y + 3$ into the first equation:

$5(y + 3) + y = 21$

Simplify:

$5y + 15 + y = 21$

$6y = 6$

$y = 1$

Now, substitute $y = 1$ back into $x = y + 3$:

$x = 1 + 3$

$x = 4$

Thus, the solution is $x = 4$, $y = 1$.

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 the system of equations:

$2x + y = 10$

$x - y = 2$

First, solve the second equation for $x$:

$x = y + 2$

Substitute $x = y + 2$ into the first equation:

$2(y + 2) + y = 10$

Expand and simplify:

$2y + 4 + y = 10$

$3y + 4 = 10$

Subtract 4 from both sides:

$3y = 6$

Divide by 3:

$y = 2$

Substitute $y = 2$ back into $x = y + 2$:

$x = 2 + 2 = 4$

Thus, the solution is:

$x = 4, y = 2$

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.