Simultaneous equations are easy to solve if we know how to graph them, but that takes a long time and sometimes we need a faster way. One efficient way to solve them is called the substitution method. As the simultaneous equations we usually work with involve $2$2 variables and $2$2 equations, this method works by solving one variable first through 'substituting' one equation into the other.
So to solve something like $3y+4=x$3y+4=x and $2+x-y=0$2+x−y=0 we'd first have to pick one equation to transform so that either $x$x or $y$y is the subject. The second equation looks more easy to manipulate, and transforming it gives us $y=2+x$y=2+x. Knowing the exact value of $y$y means we can combine the two equations by substituting $y=2+x$y=2+x into $3y+4=x$3y+4=x. This gives us:
$3\left(2+x\right)+4$3(2+x)+4 | $=$= | $x$x |
$6+3x+4$6+3x+4 | $=$= | $x$x |
$3x-x$3x−x | $=$= | $-4-6$−4−6 |
$2x$2x | $=$= | $-10$−10 |
$x$x | $=$= | $-5$−5 |
Now that we know the value of $x$x we can easily solve for $y$y by using the equation we already have that has $y$y as the subject: $y=2+x$y=2+x. Therefore $y=2+\left(-5\right)$y=2+(−5) which equals $-3$−3. So our answer is $(-5,-3)$(−5,−3).
Remember to solve for the values of both $x$x AND $y$y! Check if you have both at the end of every simultaneous question problem, unless told otherwise.
We know that the solution of a system of two simultaneous equations is represented graphically as the two lines' intersection point, or where they cross over. But did you know that simultaneous equations can also tell us whether a set of three or more lines are concurrent? When a set of lines are concurrent it means that they all cross over at the same point, like in the diagram below:
Having the same intersection point between them means that if you take any two of these three lines then their intersection point $P$P will be same as that of any other two lines.
We want to solve the following system of equations using the substitution method.
Equation 1 | $y=5x+34$y=5x+34 |
Equation 2 | $y=3x+18$y=3x+18 |
First solve for $x$x.
Hence, solve for $y$y.
Are the lines below concurrent?
Equation 1 | $-13y-2=5x$−13y−2=5x |
Equation 2 | $8=5y-x$8=5y−x |
Equation 3 | $y-3x=3$y−3x=3 |
a) First solve Equations 1 & 2 using the substitution method
Think: First look for an $x$x or $y$y term in either equation that is easy to isolate and make it the subject
Do:
The $x$x term in Equation 2 is easy to isolate:
$8$8 | $=$= | $5y-x$5y−x |
$8+x$8+x | $=$= | $5y$5y |
$x$x | $=$= | $5y-8$5y−8 |
Substituting this expression for $x$x into Equation 1:
$-13y-2$−13y−2 | $=$= | $5x$5x |
$=$= | $5\left(5y-8\right)$5(5y−8) | |
$=$= | $25y-40$25y−40 | |
$40-2$40−2 | $=$= | $25y+13y$25y+13y |
$38$38 | $=$= | $38y$38y |
$y$y | $=$= | $1$1 |
Substituting this $y$y value back into $x=5y-8$x=5y−8:
$x$x | $=$= | $5y-8$5y−8 |
$=$= | $5\times1-8$5×1−8 | |
$=$= | $5-8$5−8 | |
$=$= | $-3$−3 |
So the solution to Equations 1 & 2 is $(-3,1)$(−3,1)
b) Solve Equations 2 & 3 using the substitution method
Think: First look for an $x$x or $y$y term in either equation that is easy to isolate and make it the subject
Do:
Let's use Equation 2 again and make $x$x the subject to become $x=5y-8$x=5y−8 which we worked out in part a).
Now let's substitute it into Equation 3:
$y-3x$y−3x | $=$= | $3$3 |
$y-3\left(5y-8\right)$y−3(5y−8) | $=$= | $3$3 |
$y-15y+24$y−15y+24 | $=$= | $3$3 |
$-14y$−14y | $=$= | $-21$−21 |
$y$y | $=$= | $\frac{3}{2}$32 |
Substituting this $y$y value back into $x=5y-8$x=5y−8:
$x$x | $=$= | $5y-8$5y−8 |
$=$= | $5\times\frac{3}{2}-8$5×32−8 | |
$=$= | $\frac{15}{2}-8$152−8 | |
$=$= | $\frac{15-16}{2}$15−162 | |
$=$= | $\frac{-1}{2}$−12 |
So the solution to Equations 2 & 3 is $($($\frac{-1}{2},\frac{3}{2}$−12,32$)$)
c) Are these three lines concurrent?
Think: Concurrent lines are three or more lines that intersect at the same point. What does an intersection represent in simultaneous equations?
Do:
If the three lines are concurrent, then the intersection point between Equations 1 & 2 must be the same as Equations 2 & 3. We know that graphically the intersection point represents the solution between a system of simultaneous equations. The solution to Equations 1 & 2 is $(-3,1)$(−3,1) which is not equal to $($($\frac{-1}{2},\frac{3}{2}$−12,32$)$), the solution to Equations 2 & 3. Therefore they can not be concurrent.
We want to solve the following system of equations using the substitution method.
Equation 1 | $y=-2x-1$y=−2x−1 |
Equation 2 | $x+2y=13$x+2y=13 |
First solve for $x$x.
Hence, solve for $y$y.
We want to solve the following system of equations using the substitution method.
Equation 1 | $-7p+2q=-\frac{13}{10}$−7p+2q=−1310 |
Equation 2 | $-21p+10q=-\frac{9}{10}$−21p+10q=−910 |
First solve for $q$q.
Now solve for $p$p.
Solve systems of two linear equations involving two variables with integer coefficients, using the algebraic method of substitution or elimination