Remark
This is a very interesting question. Draw a line from the origin to where the upper right vertex of the square touches the line. That line has the property that the its equation is y = x. So the "solution" to the point of intersection is the solution of the two equations.
y = x (1)
3x + 4y = 12 (2)
Put x in for y in equation 2
3x + 4x = 12
7x = 12
x = 12/7
x = 1.714
y = 1.714
Problem A
<em><u>x intercept</u></em>
The x intercept occurs when y = 0
3x + 4(0) = 12
3x = 12 Divide by 3
x = 12/3
x = 4
the x intercept = (4,0)
<em><u>y intercept</u></em>
The y intercept occurs when x =0
3(0) + 4y = 12
4y = 12
y = 12/4
y = 3
y intercept = (0,3)
Problem B
x and y both equal 1.714 so they are also the length of the square's side.
Problem C
See solution above. x =y is the key fact.
x = y = 1.714
Since both equations are equal to y you can just make them equal to each other, like this 3x + 7 = x - 9. From there isolate the varible x on one side so that the solution will be on the other side of the equation. I'll do the steps too for you.
3x + 7 = x - 9
2x + 7 = -9
2x = -16
x = -8
Hope this helps :)
The answer is 5 since the total is 7 and you already have 2
Do cross multiply or butterfly.
For example first one.
18/x=6/10 So cross multiply 18×10 =6x
180=6x
Divide both sides by 6 to get x alone
180÷6=30
So x =30 Do the same to rest
Answer: x = 1
3x + 4 = 7
subtract 4 from both sides
3x = 3
x = 3/3 = 1
