This week I learned how to turn system word problems into equations by observing specific word that translate into math’s you first start of by declaring variables in the equation like x and y and then You search for words that mean the equal sign such as equal, is, was, will, are and most verbs. Another thing to know is that some word mean that you switch numbers like “from” an example of this is 10 from a number translates to x-10, another word that switches is “than” which can be showed through the sentence 5 is less then 10 which equals 10-5 then In general you can identify basic math terms. using all of these tools you can use them to identify and form equations and solve the systems with the different ways we have learned before. For money problems the variables we use are how many of each value like dimes or worth 0.1 dollars so the variables are how many and if we had quarter too which or worth 0.25 dollars then if we know how many we have in total which for example is 30 coins we can make and equation “d+q=30” if we know how much value of the dollars we have which is 4.20 dollars we can make the equation 0.1d+0.25=4.20 then we simply solve to find out how many of each coin we have in this situation which is 22 dimes and 8 quarters.

This week I learned how use substitution to find the solution of a system to first off you should look at one of the equations and isolate the variable that has a coefficient of either 1 or -1 to make it easier then it should look something like this 2y-3=x then in the other equation substitute the same variable you isolate into it so it would look like this 2(2y-3)+3y=6 and then isolate the y which would equal to this y=12/7 now we now that the solutions y is 12/7 we just have to input it into the original equation to find the x which is x=3/7 so the solution is          (3/7, 12/7).  you can also find the solution by simply plotting the linear equation on a graph and finding the points that match. you can also do elimination by simply multiplying either one or both equations so that either when the x or y s of both added together equals zero and the combining it altogether and isolating the non zero coefficients to get an answerhttps://myriverside.sd43.bc.ca/spencerd2021/wp-admin/post.php?post=634&action=edit#

I first started of by sectioning off parts of the graph and for the lines I made sure to use some different slopes and for each line I check where it stop and made a line according to that. what was easy was finding the domain and range for each equation. what was difficult was not make every line the same just because it looked better and find the y intercept

This week I learned that you can find the equation of a line without a graph only needing two order pairs or 1 ordered pair and the slope to do this you with the two pairs you simply find the slope of theses two lines then reduce or increase one of the order pairs by  the rise and run of the slope until the x variable is 0 and then now we know that the y number is the y-intercept to show the equation we use this where the m= the slope and the b= y-intercept and get this y=mx+b. the other way to do this is with 1 order pair and its slope we first start of by showing these number with this 1/2(x-3)=y-4. this is with the example slope 1/2 and ordered pair 3,4 we can then type this into desmos to show the line:

if we want to show this in slope intercept for we simply solve it to this: 1/2x-1.5=y-4. then we move the numbers on the y side to the x side which is this y=1/2x+2.5.

You can find slope of a line through two different ways first is by first find the y coordinate difference between one coordinate and another on the line to the right of it. so for example the y coordinate difference between (2,3) and (3,5) is 2 and then find the x difference which is 1 after that we first divide the y difference by the x difference which is 2/1 or just 2 then if needed we simplify it. Another example of this is if we have two order pair (4,2) and (5, 3,) which are points close to 1 after the other on a linear relation line the y difference is 1 and the x difference 1 so the slope of the line is 1/1 or just 1. Another way to do this to same as we do in the first way up until find the difference but instead of subtracting we instead look on a graph of the line and find how many numbers we go up by to reach the other point and the rest is the same. Good things to remember is if there is no difference in x or it is just a vertical line the slop is undefined if the line is horizontal the slop is 0.

This week in math 10 I learned how to show a function with function notation, mapping notation. To first start turning a simple equation into a function notation You write f: x as Y or in words F of x  so for example we see the function y=3x+2 in function notation we write as f(x)= 3x+2. To write a mapping notation we go from what we learned previously and change the “=” into an Arrow “->” which is shown as  f(x)->3x+2 You use mapping notation when there are a multiple amount of relation between input and output variables. Also another good way to use function notation would be to switch the F into other letters so that we can tell the difference. Another way of using function notion is in equation between different functions and showing the input an example of this is in the equation f(2)+f(8)-f(2)=8+26-8=26.

This week in math we learned how to find the X and Y intercepts of a linear equation which are where the linear relations line touches the X or Y lines this can be done in an equation to find the X you should set Y to 0 and find a number for X that works with that. an example of the X and Y intercept in a linear relationship would be with y = 2x + 1 to find the x-intercept this can be done by setting y to 0 = 2x + 1 so x would be -1/2 or -0.5 and the Y-intercept can be found by solving this y=2×0+1 which equals 1 so the X and Y intercepts are (-0.5,0) (0, 1). Another example of this being used is In this equation 3x+7y=21 so for the x-intercept we set y to 0 then we have to find out what times 3 equals 21 is 7 so the ordered pair for the x-intercept is (7,0). Now for the Y-intercept, we set x to 0 leaving us with the relation 7y=21 and we know that 7 times 3 = 21 from the other intercept so the y-intercept is (0,3).

This week I learned how to find the linear relation between two sets of numbers this can be done by first identifying the independent variable or x to input this is 1,2,3 and seeing the difference between each number in the dependent variable which is 6,8,10 which is 2 and multiplying it by the independent variable plus a number that will make the number the matching variable like 4 this can be shown like this y=2x+4. so if 1 were x that y would be 6 this is the relation between these two numbers. another example is between 1,2,3,4 and 8,4,0,-4,-8 the linear relation is -4x+12 this relation can also be plotted in a calculated to show that it Is in a straight line which means it is a linear relation.

This week I learned how to factor polynomials with three terms and determine if a trinomial is factorable. this can be done by first finding variable a which is the coefficient of the squared variable. the variable b is the variable with a power of 1 coefficient lastly the c variable is the constant. After this find, two numbers that add to b and multiply to c if there aren’t trinomial is not factorable anymore. you can show factors by having two polynomials each with one of the numbers we found and multiply those two polynomials together to check again.

An example of this is if we factor the trinomial x^2+6x+8 we find that 6 is b and 8 is C next we see the two numbers that add two 6 and multiply to 8 which are 2 and 4 because 2+4=6 and 2×4=8. we then write the factored polynomial like (x+2)(x+4). Another example of factoring trinomial can be shown with x^2+9x+8 we find that 9 is b and 8 is C next we see the two numbers that add two 1 and multiply to 8 which are 1 and 8 because 1+8=9 and 1×8=8. we then write the factored polynomial like (x+8)(x+1) we can also check to make sure that this is the right answer by solving the expression with the distributive law.

This week we learned how to multiple binomials and higher using the distributive property this can be done by first looking at the smaller polynomial and multiplying each term by each of the other polynomials’ terms individually not just the like terms and doing that with every term in the polynomial. After multiplying the terms we then combine and arrange the like terms like how we did in adding polynomials to get the simplified form of the product of both of the polynomials. I also learned a way to remember the order of the multiply the binomials with the acronym FOIL which stands for First: first terms of each binomial, Outer the outer terms of each binomial, Inner the two terms closest to the middle, Last the two last terms of each binomial. An example of the distributive property we can show with the binomials (2x-2)(4x+2) we would first start with the F which is 2x times 4x which equals 8×2 next is the O terms 2x times 2 is 4x next the I which are -2 and 4x equals -8x. Lastly, we have the L terms which are -2, and 2 which equals -4. After simplifying this polynomial we get a trinomial that is 8×2-4x-4.