This week in math 10 I learned how to solve systems of linear equations using problem solving. 

Example 1: 

The sum of two numbers is 3. Twice the larger number is 36 more than three times the smaller number.  

To solve this word problem, we will first start by reading it through once. Now during our second read through, we will declare the variables. We’ll look at the first sentence first, the sum of two numbers is 3. So, from this we can tell that two variables will need to be used that will be equal to 3 because the word is used as an equal’s sign. The equation should look like this, x+y=3. Now we’ll look at the second sentence, twice the larger number is 36 more than three times the smaller number. By looking at this sentence we can see that we need to declare a larger and smaller variable, so either x or y, for this example I will make x my larger number and y my smaller number. So now we know that 2x= 36 more than three times the smaller number. Now we need to figure out what 36 more than three times the smaller number means. Because our sentence says 36 more than three times the smaller number, we know that we must have 3y+36. Now our final equation should appear as so, 2x=3y+36. Now we need to take x+y=3 and 2x=3y+36 and solve them. We’ll solve using substitution, but to do that we need to get either variable from either equation alone. I’m going to picj the equation x+y=3 and I’m going to subtract the x from both sides of the equation, leaving me with y=-x+3. Now I’m going to input this into 2x=3y+36, so it should now look like, 2x=3(-x+3)+36. Now we’ll distribute the 3 to the –x and 3 our equation should now look like this, 2x=-3x+9+36. Now, we’ll add the 36+9=45 and we’ll take the –3x and add +3x to both sides of the equation. We know have this, 5x=45. We’re now going to divide both sides of the equation by 5, we’re left with x=9. Now to determine y we can insert x=9 into either of the original equations, y=-x+3 becomes, y=-9+3, y=-6. Our final answers are, x=9, y=-6 

Example 2: 

The perimeter of a rectangle is 40 cm. If the length were doubled and the width halved, the perimeter would be increased by 16 cm. Find the dimensions of the original rectangle. 

First, we will simply read through this equation to get it into our heads. Now we must declare the variables. The word problem states, the perimeter of a rectangle is 40 cm. Perimeter is 2L+2W and because we know is means equals sign, we can determine this equation to be, 2L+2W=40. Now we’ll look at the second sentence, If the length were doubled and the width halved, the perimeter would be increased by 16 cm. So, for this equation we must double our length, it will now be 4L, and we will halve our width, it will now be only W. Because it says the perimeter will be increased by 16cm we will now do 40+16=56, so we can write our new equation as so, 4L+W=56. Now we must take our two equations, 2L+2W=40 and 4L+W=56 and solve them. For this example, we’re going to use elimination to solve our equations. To use elimination, we need to get two of the same variables to be equal to 0 when added together, so for this example we will multiply 2L+2W=40 by –2 = -4L-4W=-80. We now need to add this to 4L+w=56. The length will be “eliminated” because it is equal to zero, the width will become –3W and the perimeter will be –24. The equation should now look like, -3W=-24. Now we will divide both sides by –3 leaving us with the equation W=8. Now we’ll insert the W=8cm into either of the original equations to find our length, 4L+8=56. We will start by subtracting 8 from both sides of the equation, 4L=48. Now we will divide both sides by 4, L=12cm. Our final measurements are, W=8cm, L=12cm.