In Problems 1-24, solve the systems algebraically.

1.
\[ x + 4y = 3 \]
\[ 4x + 2y = 9 \]

2.
\[ 3x - 2y = -5 \]
\[ 5y - 4x = 5 \]

3.
\[ 2x + 3y = 1 \]
\[ 2x - y = 1 \]

4.
\[ x + 2y = 0 \]
\[ -1 + 2y = 7 \]

5.
\[ x + y = 5 \]

6.
\[ 2p + g = 16 \]
\[ 14 - y = 7 \]
\[ 3p + 39 = 33 \]

7.
\[ x - 2y = -7 \]

8.
\[ 4x + 12y = 12 \]
\[ 5x + 3y = -9 \]

9.
\[ 2x + 4y = 12 \]
\[ 4x - 3y - 2 = 3x - 7y \]
\[ 5y - 2 = y + 4 \]
\[ 5x + 7y + 2 = 9y - 4x + 6 \]

10.
\[ 4x - y - 4 = x + y + 1 \]

11.
\[ x + 3y = 2 \]

12.
\[ z - w = 27 \]

13.
\[ 2p + 39 = 5 \]

14.
\[ 5x - 3y = 2 \]
\[ 10p + 154 = 25 \]
\[ -10x + 6y = 4 \]
\[ 2x + y + 62 = 3 \]
\[ x + y + z = -1 \]

15.
\[ x - y + 4z = 1 \]

16.
\[ 3x + y + z = 1 \]
\[ 3x + 2y - 2z = 2 \]
\[ x + 4y + 3z = 10 \]
\[ x + 2y + z = 4 \]

17.
\[ 4x + 2y - 2z = -2 \]

18.
\[ 2x - 4y - 5z = 26 \]
\[ 3x - y + z = 11 \]
\[ 2x + 3y + 2 = 10 \]

19.
\[ x - 2z = 1 \]
\[ 2y + 3z = 1 \]
\[ y + 2 = 3 \]
\[ 3x - 4z = 0 \]

20.
\[ x - y + 2z = 0 \]
\[ x - 2y - 2 = 0 \]

21.
\[ 2x + y - 2 = 0 \]

22.
\[ 2x - 4y - 2z = 0 \]
\[ x + 2y - 3z = 0 \]

23.
\[ -3y + z = 5 \]
\[ 5x + y + z = 17 \]
\[ -2x + 6y - 2z = -10 \]

24.
\[ 4x + y + z = 14 \]

25.
Mixture: A chemical manufacturer wishes to fill an order for 800 gallons of a 25% acid solution. Solutions of 20% and 35% are in stock. How many gallons of each solution must be mixed to fill the order?

26.
Mixture: A gardener has two fertilizers that contain different concentrations of nitrogen. One is 3% nitrogen and the other is 11% nitrogen. How many pounds of each should she mix to obtain 20 pounds of a 9% concentration?

To solve these systems of equations algebraically, we can use methods like substitution or elimination to eliminate one variable and solve for the other. Start by solving one equation for a variable and substitute into the other equations. Then solve the resulting equations for the variables.

Explanation:

In order to solve these systems of equations algebraically, we need to use methods like substitution or elimination to eliminate one variable and solve for the other. Let's take the first system of equations as an example:

1) x + 4y = 3 2) 4x + 2y = 9 3) 3x - 2y = -5

We can start by using the second equation to solve for one variable. Let's solve for x:

4x + 2y = 9 ⇒ 4x = 9 - 2y

Next, we can substitute this expression for x into the other two equations:

x + 4y = 3 ⇒ (9 - 2y) + 4y = 3 ⇒ 9 + 2y = 3

3x - 2y = -5 ⇒ 3(9 - 2y) - 2y = -5 ⇒ 27 - 6y - 2y = -5

Solving these resulting equations will give us the values of x and y for this system of equations. Repeat this process for each system of equations to solve them algebraically.