Linear systems using substitution

Another systematic method for solving linear systems is the substitution method. In a sense, it is very similar to the elimination method that we considered earlier, just that it implements its own elimination differently.

Given a general linear system of the form

(1)   \begin{equation*} ax+by=c \end{equation*}

(2)   \begin{equation*} dx+ey=f \end{equation*}

the substitution method proceeds as follows:

  1. isolate either x or y from equation (1) or equation (2) — choose an equation where it is easier to isolate the variable, usually an equation where the coefficient of the variable is \pm 1 or some other number that is convenient to handle;
  2. substitute the isolated variable into the other equation to obtain a single linear equation in one variable. Don’t make the mistake of substituting into the same equation from where you isolated the variable (why?);
  3. solve the resulting one-variable linear equation from step 2 above;
  4. use your answer from step 3 to obtain the value of the other variable, preferably using the isolated equation from step 1.

Example 1

Solve the linear system

(3)   \begin{equation*} x+2y=5 \end{equation*}

(4)   \begin{equation*} 7x-5y=-3 \end{equation*}

using substitution method.

We first decide on which variable to isolate. Clearly, this will be x, because it has a coefficient of 1 in equation (3). (Obviously, nothing stops us from isolating y instead, just that in this case, we’ll have to play around with the fraction it brings.)

So, from equation (3), we have

(5)   \begin{equation*} x=5-2y \end{equation*}

Next, we substitute the right side of this equation (5) (namely 5-2y) for x in equation (4):

    \[7\boxed{5-2y}-5y=-3;\]

the expression in the box being the replacement/substitute for x. We now have a linear equation in one variable, which can be easily solved:
\begin{split}
7(5-2y)-5y&=-3\\
35-14y-5y&=-3\\
35-19y&=-3\\
-19y&=-3-35\\
-19y&=-38\\
y&=\frac{-38}{-19}=2
\end{split}

It now remains to find x. Since we already isolated x as seen in equation (5), that’s the most convenient equation to use. So, putting y=2 there gives

    \[x=5-2\boxed{2}=5-2\times 2=5-4=1.\]

Finally, we check that our solution satisfies the original equations (3) and (4):

    \[\boxed{1}+2\boxed{2}=1+4=5,\quad\textrm{and}~7\boxed{1}-5\boxed{2}=7-10=-3.\]

Yes!!!

Example 2

Use substitution method to solve the linear system

(6)   \begin{equation*} 2x+3y=-7 \end{equation*}

(7)   \begin{equation*} 5x-y=-26 \end{equation*}

Since it’s easier to isolate y from equation (7), that’s what we’ll do. So, using equation (7), we have:

(8)   \begin{equation*} -y=-26-5x,\quad\implies y=26+5x \end{equation*}

Next, substitute 26+5x for y in equation (6):

    \[2x+3\boxed{26+5x}=-7\]

We’re using box to emphasize the substitution. Continuing the simplification:
\begin{split}
2x+78+15x &=-7\\
17x+78&=-7\\
17x &=-7-78\\
17x &=-85\\
x&=-\frac{85}{17}=-5
\end{split}

To find y, we use equation (8):

    \[y=26+5x=26+5\boxed{-5}=26-25=1.\]

Therefore, the solution is x=-5,~y=1.

Example 3

Solve the linear system

(9)   \begin{equation*} 3x+7y=-17 \end{equation*}

(10)   \begin{equation*} 7x-13y=19 \end{equation*}

using substitution method.

Which variable should we first isolate in this case? The answer is either x or y. Whichever one is chosen, we’ll still have to deal with the fractions that come with it. Actually, we should expect to come across fractions in our linear-systems-by-substitution-journey (or more generally, in our arithmetical journey).

Say we decide to isolate x from equation (9):

(11)   \begin{equation*} x=\frac{-17-7y}{3} \end{equation*}

Then we substitute the right side of equation (11) for x in equation (10), obtaining
\begin{split}
7\boxed{\frac{-17-7y}{3}}-13y&=19\\
\boxed{\frac{-17\times 7-7y\times 7}{3}}-13y &=19\\
\boxed{\frac{-119-49y}{3}}-13y &=19\\
3\Big(\frac{-119-49y}{3}\Big)-3(13y) &=3(19)\\
-119-49y-39y &=57\\
-49y-39y &=57+119\\
-88y &=176\\
y &=\frac{176}{-88}=-2
\end{split}

To get x we use equation (11):

    \[x=\frac{-17-7y}{3}=\frac{-17-7(-2)}{3}=\frac{-17+14}{3}=\frac{-3}{3}=-1.\]

Thus, the solution to our linear system is x=-1, y=-2.

Example 4

Find the point of intersection of the lines

(12)   \begin{equation*} x-2y=3 \end{equation*}

(13)   \begin{equation*} 3x-6y=4 \end{equation*}

using substitution method.

From equation (12) we isolate x:

(14)   \begin{equation*} x=3+2y \end{equation*}

and then substitute the right side of equation (14) for x in equation (13):
\begin{split}
3\boxed{3+2y}-6y &=4\\
9+6y-6y &=4\\
9&=4.
\end{split}
Since 9\neq 4, the last line above is a contradiction. Since we arrived at a contradiction, the system is inconsistent and so the two lines do not intersect.

Example 5

Use substitution to solve the linear system

(15)   \begin{equation*} 3x-2y=23 \end{equation*}

(16)   \begin{equation*} 21x-14y=161 \end{equation*}

We begin by isolating x from equation (15):

(17)   \begin{equation*} x=\frac{2y+23}{3} \end{equation*}

Then we substitute the right side of equation (17) for x in equation (16):
\begin{split}
21\boxed{\frac{2y+23}{3}}-14y&=161\\
7(2y+23)-14y&=161\\
14y+161-14y&=161\\
161&=161,
\end{split}
which is always true. Since we obtained an identity, the linear system has infinite number of solutions. The two lines coincide.

Example 6

Divide 32 into two parts such that one part is 25\% of the other.

Let’s apply substitution technique to solve the above problem, although it can be solved easily by a different means. So, let the two parts be x and y. We have the following linear system

(18)   \begin{equation*} x+y=32 \end{equation*}

(19)   \begin{equation*} x=25\% y, \quad\implies x=\frac{1}{4}y,\quad\implies y=4x \end{equation*}

Substituting y=4x in equation (18) gives x+4x=32, and so x=\frac{32}{5}=6.4. Then y=4\times \frac{32}{5}=\frac{128}{5}=25.6. Therefore, the two parts are 6.4 and 25.6. (Check that 6.4+25.6=32 and that 6.4=\frac{1}{4}\times 25.6.)

Example 7

Fifty six coins, comprising entirely of nickels and quarters, add up to six dollars. Determine the number of nickels and the number of quarters.

Sometimes it is useful to choose variables that are somewhat related to what we want. So, let the number of nickels be n and let the number of quarters be q. Since there are fifty six coins, we have n+q=56. Since a nickel is 5 cents and a quarter is 25 cents, we have that 5n+25q=600. Notice how we changed six dollars to six hundred cents, for convenience.

(20)   \begin{equation*} n+q=56 \end{equation*}

(21)   \begin{equation*} 5n+25q=600,\quad\implies n+5q=120 \end{equation*}

Isolate n from equation (20):

(22)   \begin{equation*} n=56-q \end{equation*}

Substitute the right side of equation (22) for n in equation (21):
\begin{split}
\boxed{56-q}+5q&=120\\
56+4q&=120\\
4q&=120-56\\
4q&=64\\
q&=16\\
&\vdots\vdots\vdots\vdots\vdots\\
n&=56-q\\
&=56-16\\
&=40
\end{split}
Since we obtained n=40 and q=16, it follows that there are 40 nickels and 16 quarters. Check that 40+16=56 and that 40\times 0.05 + 16\times 0.25=2+4=6.

Digression: Using substitution to balance chemical equations

You may have come across balancing chemical equations if you’ve taken a course on chemistry before. The procedure is usually by trial-and-error, but it is also possible to use linear systems especially for more complex chemical equations where the trial-and-error method is inconvenient.

Example 8

Balance the chemical equation

(23)   \begin{equation*} \ce{KClO3 -> KCl + O2} \end{equation*}

Obviously, we can easily balance the above equation by trial-and-error. We’ve chosen such a simple example for the purpose of illustration.

Begin by assigning arbitrary coefficients to the reactants and products:

    \[\ce{xKClO3 -> yKCl + zO2}\]

and then compare coefficients of like atoms.
\begin{split}
\textrm{Potassium, K}: x&=y\\
\textrm{Chlorine, Cl}: x&=y\\
\textrm{Oxygen, O}:3x&=2z
\end{split}
From the last equation, we have z=\frac{3}{2}x. Since z should be an integer, x has to be a multiple of 2. In particular, x=2, because we have to choose the minimum possible value. In turn, z=\frac{3}{2}2=3. And then, y=2 because x=y. Putting these values into equation (23) we obtain the balanced equation

    \[\ce{2KClO3 -> 2KCl + 3O2}\]

Example 9

Balance the chemical equation

(24)   \begin{equation*} \ce{Na + H2O -> NaOH + H2} \end{equation*}

As before, we assign arbitrary coefficients to the reactants and products:

    \[\ce{xNa + yH2O -> zNaOH + wH2}\]

and then equate the coefficients of like atoms:
\begin{split}
\textrm{Sodium, Na}: x&=z\\
\textrm{Hydrogen, H}: 2y&=z + 2w\\
\textrm{Oxygen, O}: y&=z\\
&\vdots\vdots\vdots\vdots\vdots\\
2y&=y+2w ~(\textrm{putting} ~y=z~\textrm{in}~2y=z+2w)\\
2y-y&=2w\\
y&=2w\\
&\vdots\vdots\vdots\vdots\vdots\\
z&=2w ~(\textrm{because} ~y=z)\\
x&=2w ~(\textrm{because} ~x=z)
\end{split}
So we have y=2w,~x=2w,~z=2w. Whatever positive integer value is assigned to w, the values of x,y,z will all be integers. We choose the minimum possible value of w, namely 1.
\begin{split}
w&=1\\
x&=2w=2\times 1=2\\
y&=2w=2\times 1=2\\
z&=2w=2\times 1=2
\end{split}
Substituting these values in equation (24), we obtain

    \[\ce{2Na + 2H2O -> 2NaOH + H2,}\]

which is now balanced.

Example 10

Balance the chemical equation

(25)   \begin{equation*} \ce{NaHCO3 -> Na2CO3 + H2O + CO2} \end{equation*}

Begin by assigning arbitrary coefficients to the reactants and products:

    \[\ce{xNaHCO3 -> yNa2CO3 + zH2O + wCO2}\]

and then compare coefficients of like atoms:
\begin{split}
\textrm{Sodium, Na}: x&=2y\\
\textrm{Hydrogen, H}: x&=2z\quad\textrm{so}~2y=2z~\textrm{or}~y=z\\
\textrm{Carbon, C}: x&=y+w\quad\textrm{so}~2y=y+w~\textrm{or}~2y-y=w\\
\textrm{Oxygen, O}: 3x&=3y+z+2w\\
&\vdots\vdots\vdots\vdots\vdots\vdots\vdots\vdots\vdots\vdots\\
x&=2y,\quad y=z,\quad y=w
\end{split}
We choose y=1 and obtain
\begin{split}
x&=2y=2\times 1=2\\
z&=y=1\\
w&=y=1
\end{split}
Returning to equation (25), the balanced equation is

    \[\ce{2NaHCO3 -> Na2CO3 + H2O + CO2}.\]

Don’t worry about the last three examples, especially if you’re still in grade 10 and won’t take chemistry in future.

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