PLEASE HELP due 1-22-21 Solving Functions with substitution and elimination
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HallsofIvy
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You say "how do i solve the equation" but NONE of these problems asks you to solve an equation!
The first four ask you to solve a pair of equations by graphing them. Do you know what that means? These are all "linear equations" which means their graphs are straight lines. And straight lines are determined by two points.
The first problem has \(\displaystyle y= -\frac{1}{4}x- 4\) and \(\displaystyle y= \frac{5}{4}x+ 2\).
When x= 0, the first equation gives \(\displaystyle y=0 -4= -4\) so one point is (0, -4).
When x= 12 (a multiple of 4 to cancel the denominator and avoid fractions), the first equation gives \(\displaystyle y= -3- 4= -7\) so another point is (12, -7). Mark those two points on the graph and draw the line through them. That line is the graph of \(\displaystyle y=\frac{1}{4}x- 4\).
When x= 0, the second equation gives \(\displaystyle y= 0+ 2= 2\) so one point is (0, 2).
When x= 12, the second equation gives \(\displaystyle y= 5+ 2= 7\) so another point is (12, 7). Mark those two points on the graph and draw the line between them. That line is the graph of \(\displaystyle y= \frac{5}{4}x+ 2\).
The point of where those lines cross is the (x, y) that satisfies both equations.
Exercises 2, 3, and 4 are done the same way.
Exercise 5 has the two equations x+ 8y= -15 and 7x+ 8y= -9 and you are to solve them by "substitution". Do you know what THAT means? Subtracting 8y from both sides of the first equation gives x= -8y- 15. SUBSTITUTE that for the "x" in the second equation to get 7(-8y- 15)+ 8y= -56y- 105+ 8y= -48y- 105= -9. Can you solve that for y? Once you get y, put that value into x= -8y- 15.
The first four ask you to solve a pair of equations by graphing them. Do you know what that means? These are all "linear equations" which means their graphs are straight lines. And straight lines are determined by two points.
The first problem has \(\displaystyle y= -\frac{1}{4}x- 4\) and \(\displaystyle y= \frac{5}{4}x+ 2\).
When x= 0, the first equation gives \(\displaystyle y=0 -4= -4\) so one point is (0, -4).
When x= 12 (a multiple of 4 to cancel the denominator and avoid fractions), the first equation gives \(\displaystyle y= -3- 4= -7\) so another point is (12, -7). Mark those two points on the graph and draw the line through them. That line is the graph of \(\displaystyle y=\frac{1}{4}x- 4\).
When x= 0, the second equation gives \(\displaystyle y= 0+ 2= 2\) so one point is (0, 2).
When x= 12, the second equation gives \(\displaystyle y= 5+ 2= 7\) so another point is (12, 7). Mark those two points on the graph and draw the line between them. That line is the graph of \(\displaystyle y= \frac{5}{4}x+ 2\).
The point of where those lines cross is the (x, y) that satisfies both equations.
Exercises 2, 3, and 4 are done the same way.
Exercise 5 has the two equations x+ 8y= -15 and 7x+ 8y= -9 and you are to solve them by "substitution". Do you know what THAT means? Subtracting 8y from both sides of the first equation gives x= -8y- 15. SUBSTITUTE that for the "x" in the second equation to get 7(-8y- 15)+ 8y= -56y- 105+ 8y= -48y- 105= -9. Can you solve that for y? Once you get y, put that value into x= -8y- 15.