Solve the following system of equations by graphing.

[latex]\begin {cases} 2x+y=-8\\x-y=-1\end {cases}[/latex]

Solve the first equation for [latex]y.[/latex]

[latex]\begin{array}{rcl}2x+y&=&-8\\y&=&-2x-8&\end{array}[/latex]

Solve the second equation for [latex]y.[/latex]

[latex]\begin{array}{rcl}x-y&=&-1\\y&=&x+1\end{array}[/latex]

Graph both equations on the same set of axes.

The lines appear to intersect at the point [latex](-3, -2).[/latex]

We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.

[latex]\begin{array}{rclr}2(-3)+(-2) & = & -8 \\ -8 & = & -8 &\text{True}\\ \\(-3)-(-2) & =&-1\\-1&=&-1&\text{True}\end{array}[/latex]

The solution to the system is the ordered pair [latex](-3, -2)[/latex].

Julia has just retired and has $600,000 in her retirement account that she needs to reallocate to produce income. She is looking at two investments: a very safe guaranteed annuity that will provide 3% interest and a somewhat riskier bond fund that averages 7% interest. She would like to invest as little as possible in the riskier bond fund but needs to produce $40,000 a year in interest to live on. How much should she invest in each account?

Notice there are two unknowns in this problem: the amount she should invest in the annuity and the amount she should invest in the bond fund. We can start by defining variables for the unknowns:

[latex]a:[/latex] The amount (in dollars) she invests in the annuity

[latex]b:[/latex] The amount (in dollars) she invests in the bond fund

Our first equation comes from noting that together she is going to invest $600,000: [latex]a+b=600,000[/latex]

Our second equation will come from the interest. She earns 3% on the annuity, so the interest earned in a year would be [latex]0.03a.[/latex] Likewise, the interest earned on the bond fund in a year would be [latex]0.07b.[/latex] Together, these need to total $40,000, giving the equation: [latex]0.03a+0.07b=40,000[/latex]

Together, these two equations form our system. The first equation is an ideal candidate for the first step of substitution—we can easily solve the equation for [latex]a[/latex] or [latex]b:[/latex] [latex]a=600,000-b[/latex]

Then we can substitute this expression for [latex]a[/latex] in the second equation and solve.

[latex]\begin {array}{rcl}0.03(600,000-b)+0.07b&=&40,000\\18,000-0.03b+0.07b&=&40,000\\0.04b&=&22,000\\b&=&550,000\end {array}[/latex]

Now substitute this back into the equation [latex]a=600,000-b[/latex] to find [latex]a[/latex].

[latex]\begin{array}{rcl}a&=&600,000-550,000\\a&=&50,000\end{array}[/latex]

In order to reach her goal, Julia will have to invest $550,000 in the bond fund and $50,000 in the annuity.

Solve the following system by elimination.

[latex]\begin{cases}20b+30p=3900\\15b+30p=3300\end{cases}[/latex]

Adding the equations would not eliminate a variable, but we notice that the coefficients on [latex]p[/latex] are the same, so multiplying one of the equations by –1 will change the sign of the coefficients. Multiplying the second equation by –1 gives the system

[latex]\begin{array}{rcl}20b+30p&=&3900\\-15b-30p&=&-3300\end{array}[/latex]

Adding these equations gives

[latex]\begin{array}{rcl}5b&=&600\\b&=&120\end{array}[/latex]

Substituting [latex]b = 30[/latex] into the first equation,

[latex]\begin{array}{rcl}20(120)+30p&=&3900\\2400+30p&=&3900\\30p&=&1500\\p&=&50\end{array}[/latex]

The solution is [latex]b = 120[/latex], [latex]p = 50[/latex].

Checking our answer in the second equation:

[latex]\begin{array}{rcl}15(120)+30(50)&=&3300\\1800+1500&=&3300\\3300&=&3300\end{array}[/latex]

Solve the given system of equations in two variables by addition.

[latex]\begin{cases}2x+3y=-16\\5x+10y=30\end{cases}[/latex]

One equation has [latex]2x[/latex] and the other has [latex]5x[/latex]. The least common multiple is [latex]10x[/latex], so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate [latex]x[/latex] by multiplying the first equation by [latex]-5[/latex] and the second equation by [latex]2[/latex].

[latex]\begin{array}{rcl}-5(2x+3y)&=&-5(-16)\\-10x-15y&=&80\\2(5x-10y)&=&2(30)\\10x-20y&=&60\end{array}[/latex]

Then we add the two equations together.

[latex]\begin{array}{rcl}-10x-15y&=&80\\10x-20y&=&60\\ \hline-35y&=&140\\y&=&-4\end{array}[/latex]

Substitute [latex]y=-4[/latex] into the original first equation.

[latex]\begin{array}{rcl}2x+3(-4)&=&-16\\2x-12&=&-16\\2x&=&-4\\x&=&-2\end{array}[/latex]

The solution is [latex](-2, -4)[/latex]. Check it in the other equation.

[latex]\begin{array}{rcl}5x-10y&=&30\\5(-2)-10(-4)&=&30\\-10+40&=&30\\30&=&30\end{array}[/latex]

Find a solution to the system of equations using the addition method.

[latex]\begin{cases}x+3y=2\\3x+9y=6\end{cases}[/latex]

 With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating [latex]x[/latex]. If we multiply both sides of the first equation by –3, then we will be able to eliminate the x-variable.

[latex]\begin{array}{rclc}x+3y&=&2\\(-3)(x+3y)&=&(-3)(2)&\text{Multiply both sides of the equation by -3.}\\-3x-9y&=&-6\end {array}[/latex]

Now add the equations together to eliminate the [latex]x[/latex] variable.

[latex]\begin{array}{rcl}-3x-9y&=&-6\\3x+9y&=&6\\ \hline 0&=&0\end{array}[/latex]

We can see that there will be an infinite number of solutions that satisfy both equations. In some cases, realizing there are an infinite number of solutions is enough, and we can stop there. In other cases, we will want to describe the set of solutions.

Determine whether the ordered triple [latex](3, -2, 1)[/latex] is a solution to the system.

[latex]\begin{array}{rclc}x+y+z&=&2&(1)\\6x-4y+5z&=&31&(2)\\5x+2y+2z&=&13&(3)\end{array}[/latex]

We will check each equation by substituting in the values of the ordered triple for [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex].

Equation (1):

[latex]\begin{array}{rclc}x+y+z&=&2\\(3)+(-2)+(1)&=&2&\text{True}\end{array}[/latex]

Equation (2):

[latex]\begin{array}{rclc}6x-4y+5z&=&31\\ 6(3)-4(-2)+5(1)&=&31\\ 18+8+5&=&31&\text{True}\end{array}[/latex]

Equation (3):

[latex]\begin{array}{rclc}5x+2y+2z&=&13\\ 5(3)-2(-2)+2(1)&=&13\\15-4+2&=&13&\text{True}\end{array}[/latex]

The ordered triple [latex](3, -2, 1)[/latex] is indeed a solution to the system.

Find a solution to the following system. The equations are numbered so we can refer to them more easily.

[latex]\begin{cases} x-2y+3z=9 \text{ } (1)\\-x+3y-z=-6 \text{ } (2)\\2x-5y+5z=17 \text{ } (3)\end{cases}[/latex]

There will always be several choices as to where to begin, but the most obvious first step here is to eliminate [latex]x[/latex] by adding equations (1) and (2).

[latex]\begin{array}{rclc}x-2y+3z&=&9&(1)\\-x+3y-z&=&-6&(2)\\ \hline y+2z&=&3&(4)\end{array}[/latex]

The second step is multiplying equation (1) by –2 and adding the result to equation (3). These two steps will eliminate the variable [latex]x[/latex].

[latex]\begin{array}{rclcc}-2x+4y-6z&=&-18&(1)&\text{multiplied by -2}\\2x-5y+5z&=&17&(3)\\ \hline -y-z&=&-1&(5)\end{array}[/latex]

In equations (4) and (5), we have created a new two-by-two system. We can solve for [latex]z[/latex] by adding the two equations.

[latex]\begin{array}{rclc}y+2z&=&3&(4)\\-y-z&=&-1&(5)\\ \hline z&=&2&(6)\end{array}[/latex]

Choosing one equation from each new system, we obtain the upper triangular form:

[latex]\begin{array}{rclc}x-2y+3z&=&9&(1)\\y+2z&=&3&(4)\\z&=&2&(6)\end{array}[/latex]

Next, we back-substitute [latex]z=2[/latex] into equation (4) and solve for [latex]y[/latex].

[latex]\begin{array}{rcl}y+2(2)&=&3\\y+4&=&3\\y&=&-1\end{array}[/latex]

Finally, we can back-substitute [latex]z=2[/latex] and [latex]y=-1[/latex] into equation (1). This will yield the solution for

[latex]\begin{array}{rcl}x-2(-1)+3(2)&=&9\\x+2+6&=&9\\x&=&1\end{array}[/latex]

The solution is the ordered triple [latex](1, -1, 2)[/latex].

Chad is trying to plan a meal to meet specific nutritional goals. He wants to construct a dish containing rice, tofu, and peanuts that will provide 30g of protein, 14g of fat, and 50g of carbohydrates. How much of each ingredient should he use?

First off, we’re assuming any other ingredients used in the recipe aren’t contributing significantly enough to the nutrition to be considered. To answer this question, we’ll first need to know the nutritional content for the ingredients. Looking these up:

White rice: 1 cup provides: 0g fat, 44g carbohydrates, 4g protein

Tofu: 1 cup provides: 10g fat, 5g carbohydrates, 20g protein

Peanuts: 1 cup provides: 72g fat, 31g carbohydrates, 35g protein

Now we can define our variables. We are interested in the amount of each ingredient to use, so we’ll define our variables as the quantity of each ingredient:

[latex]r[/latex]: cups of rice, [latex]t[/latex]: cups of tofu, [latex]p[/latex]: cups of peanuts.

Now for each nutrient, we can create an equation. Since 1 cup of rice provides [latex]44g[/latex] carbohydrates, [latex]r[/latex] cups will provide [latex]44r[/latex] grams of carbohydrates. Likewise [latex]t[/latex] cups of tofu will provide [latex]5t[/latex] grams, and [latex]p[/latex] cups of peanuts will provide [latex]31p[/latex] grams. Together we want our recipe to provide [latex]50g[/latex] of carbohydrates, giving the equation:

[latex]44r+5t+31p=50[/latex]

Doing the same for fat and protein gives the full system:

[latex]\begin{cases}44r+5t+31p=50\\10t+72p=14\\4r+20t+35p=30\end{cases}[/latex]

Now we can solve the system.

Step 1. Notice that the section equation already does not involve the variable [latex]r[/latex]. To make things simpler, a first step might be to interchange the last two equations so the two equations with three variables will line up.

[latex]\begin{array}{rclc}44r+5t+31p&=&50&(1)\\4r+20t+35p&=&30&(2)\\10t+72p&=&14&(3)\end{array}[/latex]

Step 2. Since [latex]r[/latex] is already eliminated in the last equation, we’ll eliminate [latex]r[/latex] from the first two equations. Multiply equation (2) by –11.

[latex]\begin{array}{rclc}44r+5t+31p&=&50&(1)\\-44r-220t-385p&=&-330&(2)\\10t+72p&=&14&(3)\end{array}[/latex]

Step 3. Add equations (1) and (2), writing the result as row 2.

[latex]\begin{array}{rclc}44r+5t+31p&=&50&(1)\\-215t-354p&=&-280&(2)\\10t+72p&=&14&(3)\end{array}[/latex]

Step 4. Multiply equation (2) by 2 and equation (3) by 43.

[latex]\begin{array}{rclc}44r+5t+31p&=&50&(1)\\-430t-708p&=&-560&(2)\\430t+3096p&=&602&(3)\end{array}[/latex]

Step 5. Add equations (2) and (3), writing the result in row 3.

[latex]\begin{array}{rclc}44r+5t+31p&=&50&(1)\\-430t-708p&=&-560&(2)\\2388p&=&42&(3)\end{array}[/latex]

Step 6. Solve for [latex]p[/latex] in equation 3. For a real-life problem like this, decimal approximations are probably fine.

[latex]\begin{array}{rcl}2388p&=&42\\p&=&\frac{42}{2388}\approx{0.0176}\end{array}[/latex]

Step 7. Back-substitute the value for [latex]p[/latex] into equation (2) to solve for [latex]t[/latex].

[latex]\begin{array}{rcl}-430t-708(0.0176)&=&-560\\-430t-12.4608&=&-560\\-430t&=&-547.5392\\t&=&\frac{-547.5392}{-430}\approx{1.273}\end{array}[/latex]

Step 8. Back-substitute the values for [latex]p[/latex] and [latex]t[/latex] into equation (1) and solve for [latex]r[/latex].

[latex]\begin{array}{rcl}44r+5(1.273)+31(0.0176)&=&50\\44r&=&43.0894\\r&\approx{ }&0.979\end{array}[/latex]

To meet his nutritional goals, Chad should use 0.979 cups of rice, 1.273 cups of tofu, and 0.0176 cups of peanuts.