Question

How do you solve the following linear system? :$7x + 3y = 27,3x - 4y = 7$?

Answer 1

Hint: Make the coefficient of any one of the variables the same by multiplying the equations with respective constants.
Add or subtract them to eliminate that variable from the equation.
Find the value of one variable from the new equation.
Substitute the value of this variable in one of the two original equations to get the second variable.

Complete step by step solution:
The given equation is $7x + 3y = 27,3x - 4y = 7$
$7x + 3y = 27 \to (1)$
$3x - 4y = 7 \to (2)$
We use the elimination method
Multiply first equation by $4$ and second by $3$, hence we get
$(1) \times 4 \Rightarrow 28x + 12y = 108$
$(2) \times 3 \Rightarrow 9x - 12y = 21$
Now we use the elimination method
Add these two equations to eliminate $y$
$
(28x+12y=108)+(9x-12y=21)
\\ \Rightarrow 28x+9x=12y-12y=108+21
$
The zero terms vanish, hence we get
$\Rightarrow$$37x = 129$
Divide by $37$ on both sides, hence we get
$\Rightarrow$$\dfrac{{\not{{37}}}}{{\not{{37}}}}x = \dfrac{{129}}{{37}}$
$\Rightarrow$$x = \dfrac{{129}}{{37}}$
Now substitute this value into the second equation, hence we get
$3x - 4y = 7$
$\Rightarrow$$3 \times \dfrac{{129}}{{37}} - 4y = 7$
Interchange the variable and constant, hence we get
$\Rightarrow$$4y = 3 \times \dfrac{{129}}{{37}} - 7$
Now simplify the RHS (Right Hand Side)
$\Rightarrow$$4y = \dfrac{{387 - 259}}{{37}}$
Subtract the numerator, hence we get
$\Rightarrow$$y = \dfrac{{128}}{{4 \times 37}} = \dfrac{{128}}{{148}}$
The result is
$\Rightarrow$$y = \dfrac{{64}}{{74}} = \dfrac{{32}}{{37}}$

Therefore the value of x is $\dfrac{{129}}{{37}}$ and the value of y is $\dfrac{{32}}{{37}}$.

Note: We use the alternative method.
The alternative method is the substitution method.
The given equation is $7x + 3y = 27,3x - 4y = 7$
$7x + 3y = 27 \to (1)$
$3x - 4y = 7 \to (2)$
We use the substitution method.
Let’s take the first equation.
$7x + 3y = 27$
Isolate the $x$ term, hence we get
$\Rightarrow$\[7x = 27 - 3y\]
Divide by $7$ on both sides
$\Rightarrow$\[\dfrac{{\not{7}}}{{\not{7}}}x = \dfrac{{27 - 3y}}{7}\]
Then,
\[x = \dfrac{{27 - 3y}}{7}\]
Now, the $x$ value substitute in the second equation. The second equation is $3x - 4y = 7$, hence we get
$\Rightarrow$$3\left( {\dfrac{{27 - 3y}}{7}} \right) - 4y = 7$
Now we simplify this equation
Multiply by $3$
$\Rightarrow$$\left( {\dfrac{{81 - 9y}}{7}} \right) - 4y = 7$
Now take LCM in LHS (Left Hand Side), hence we get
$\Rightarrow$\[\dfrac{{81 - 9y - 4y \times 7}}{7} = 7\]
Multiply $ - 4y$ by $7$, hence we get
$\Rightarrow$\[\dfrac{{81 - 9y - 28y}}{7} = 7\]
Add the $y$ variable in the numerator, hence we get
$\Rightarrow$\[\dfrac{{81 - 37y}}{7} = 7\]
Multiply by $7$ on both sides, hence we get
$\Rightarrow$$81 - 37y = 49$
Now we isolate the $y$
$\Rightarrow$$37y = 81 - 49$
Subtract on RHS
$\Rightarrow$$37y = 32$
Divide by $37$ on both sides
$\Rightarrow$$y = \dfrac{{32}}{{37}}$
Now we find the $y$ value
The $y$ value substitute in the second equation
$\Rightarrow$$3x - 4\left( {\dfrac{{32}}{{37}}} \right) = 7$
Multiply by $4$ in the numerator, hence we get
$\Rightarrow$$3x - \dfrac{{128}}{{37}} = 7$
Now take LCM in LHS (Left Hand Side), hence we get
$\Rightarrow$$\dfrac{{3x(37) - 128}}{{37}} = 7$
Multiply $3x$ by $37$, hence we get
$\Rightarrow$ $\dfrac{{111x - 128}}{{37}} = 7$
Multiply by $37$ on both sides, hence we get
$\Rightarrow$$111x - 128 = 259$
Isolate the $x$ term, hence we get
$\Rightarrow$$111x = 259 + 128$
Add the denominator
$\Rightarrow$ $111x = 387$
Divide by$111$on both sides, hence we get
$\Rightarrow$$x = \dfrac{{387}}{{111}}$
Now we divide by $111$
$\Rightarrow$$x = \dfrac{{129}}{{37}}$
Hence, $x = \dfrac{{129}}{{37}}$ and $y = \dfrac{{32}}{{37}}$.