ORCCAOpen Resources for Community College Algebra

The first equation, \(y=\frac{3}{2}(x-1)+4\text{,}\) is a linear equation in point-slope form with a slope of \(\frac{3}{2}\) that passes through the point \((1,4)\text{.}\) The second equation, \(3x-2y=4\text{,}\) is a linear equation in standard form To graph this line, we either need to find the intercepts or put the equation into slope-intercept form. Just for practice, we will put the line in slope-intercept form.

We'll use this information to graph both lines:

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The two lines never intersect: they are parallel. So there are no solutions to the system of equations. We write the solution set as \(\emptyset\text{.}\)

We need to solve for one of the variables in one of our equations. Looking at both equations, it will be best to solve for \(y\) in the second equation. The coefficient of \(y\) in that equation is smallest.

To check this solution, we replace \(\substitute{x}\) with \(\substitute{-4}\) and \(\lighthigh{y}\) with \(\lighthigh{-5}\) in each equation:

We conclude then that this system of equations is true when \(x=-4\) and \(y=-5\text{.}\) Our solution is the point \((-4,-5)\) and we write the solution set as \(\{(-4,-5)\}\text{.}\)

Let's let \(a\) represent the number of adult tickets sold and \(c\) represent the number of child tickets sold. We need to build two equations to solve a system for both variables.

The first equation we will build relates to the fact that there were \(79\) total tickets sold. If we combine both the number of adult tickets and child tickets, the total is \(79\text{.}\) This fact becomes:

For the second equation we need to use the per-ticket dollar amounts to generate the total cost of \(\$384\text{.}\) The amount of money that was made from adult tickets is found my multiplying the number of adult tickets sold, \(a\text{,}\) by the price per ticket, \(\$6\text{.}\) Similarly, the amount of money from child tickets is found my multiplying the number of child tickets sold, \(c\text{,}\) by the price per ticket, \(\$4\text{.}\) These two amounts will add to be \(\$384\text{.}\) This fact becomes:

And so, our system is

To solve, we will use the substitution method and solve the first equation for the variable \(a\text{.}\)

Now we will substitute \(\substitute{79-c}\) for \(\substitute{a}\) in the second equation.

Last, we will solve for \(a\) by substituting \(\substitute{45}\) in for \(\substitute{c}\) in the equation \(a=79-c\text{.}\)

Our conclusion is that Amanda sold \(34\) adult tickets and \(45\) child tickets.

To solve the systems using substitution, we first need to solve for one variable in one equation, then substitute into the other equation.

1. In this case, \(x\) is already solved for in the second equation so we can substitute \(\substitute{\frac{5}{3}y+3}\) everywhere we see \(\substitute{x}\) in the first equation. Then simplify and solve for \(y\text{.}\)

   \begin{align*} 3\substitute{x}-5y\amp=9\\ 3\left(\substitute{\frac{5}{3}y+3}\right)-5y\amp=9\\ 5y+9-5y\amp=9\\ 9\amp=9 \end{align*}

   We will stop here since we have eliminated all of the variables in the equation and ended with a true statement. Since \(9\) always equals \(9\text{,}\) no matter what, then any value of \(y\) must make the original equation, \(3\left(\frac{5}{3}y+3\right)-5y=9\) true. If you recall from the section on substitution, this means that both lines \(3x-5y=9\) and \(x=\frac{5}{3}y+3\) are in fact the same line. Since a solution to a system of linear equations is any point where the lines touch, all points along both lines are solutions. We can say this in English as, “The solutions are all points on the line \(3x-5y=9\text{,}\)” or in math as, “The solution set is \(\{(x,y)|3x-5y=9\}\text{.}\)”

2. We will first solve the top equation for \(y\text{.}\)

   \begin{align*} y+7\amp=4x\\ y\amp=4x-7 \end{align*}

   Now we can substitute \(\substitute{4x-7}\) wherever we see \(\substitute{y}\) in the second equation.

   \begin{align*} 2\substitute{y}-8x\amp=7\\ 2\left(\substitute{4x-7}\right)-8x\amp=7\\ 8x-14-8x\amp=7\\ -14\amp=7 \end{align*}

   We will stop here since we have eliminated all of the variables in the equation and ended with a false statement. Since \(-14\) never equals \(7\text{,}\) then no values of \(x\) and \(y\) can make the original system true. If you recall from the section on substitution, this means that the lines \(y+7=4x\) and \(2y-8x=7\) are parallel. Since a solution to a system of linear equations is any point where the lines touch, and parallel lines never touch, no points are solutions. We can say this in English as, “There are no solutions,” or in math as, “The solution set is \(\emptyset\text{.}\)”

To solve the system using elimination, we first need to scale one or both of the equations so that one variable has equal but opposite coefficients in the system. In this case, we will choose to make \(y\) have opposite coefficients because the signs are already opposite for that variable in the system.

We need to multiply the first equation by \(2\) and the second equation by \(3\text{.}\)

We now have an equivalent system of equations where the \(y\)-terms can be eliminated:

To solve for \(y\text{,}\) we will substitute \(\substitute{1}\) for \(\substitute{x}\) into either of the original equations. We will use the first equation, \(4x-6y=13\text{:}\)

To verify this, we substitute the \(x\) and \(y\) values into both of the original equations.

So the solution is the point \(\left(-\frac{3}{2},1\right)\) and the solution set is \(\left\{\left(-\frac{3}{2},1\right)\right\}\text{.}\)

To eliminate the \(x\)-terms, we will scale the second equation by \(-3\text{.}\)

Adding the respective sides of the equation, we have:

Both of the variables have been eliminated. In this case, the statement \(0=3\) is just false, no matter what \(x\) and \(y\) are. So the system has no solution. The solution set is \(\emptyset\text{.}\)

1. Elimination is probably easiest here. Multiply the first equation by \(2\) and eliminate the \(y\) variables. The solution to this one is \((-1,-3)\) if you want to solve it for practice.

2. Substitution is probably easiest here. Solve the first equation for \(x\) and substitute it into the second equation. We could use elimination if we multiplied the first equation by \(-9\) and eliminate the \(x\) variable, but it's probably a little more work than substitution. The solution to this one is \((-4,-2)\) if you want to solve it for practice.

3. Elimination is probably easiest here. Multiply the first equation by \(2\) and the second equation by \(-3\text{.}\) Doing this will eliminate both variables and leave you with \(0=0\text{.}\) This should mean that all points on the line are solutions. So the solution set is \(\{(x,y)|6x+30y=15\}\text{.}\)

4. Substitution is definitely easiest here. Substituting \(y\) from one equation into \(y\) in the other equation gives you \(3x-2=7x+6\text{.}\) Solve this and find then find \(y\) and you should get the solution to the system to be \((-2,-8)\) if you want to solve it for practice.