Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions
Page 95 Exercise 1 Answer
Given: The graph with points plotted in it
To find the Ordered pair of A
Solution: From the graph, we can see the cartesian coordinator
Firstly we have to check where A lies in the graph and then draw a horizontal perpendicular line and vertical perpendicular lines to the y-axis and x-axis respectively.
From the graph, we can clearly see the location of point A, the x-axis to the origin is, 2, and the y-axis to the origin is 6. The ordered pair is written as which is (2,6)
The ordered pair of A is (2,6)
Page 95 Exercise 3 Answer
Given: Graph with different points on it
To find the ordered pair of E
Solution: The cartesian plotting from the graph is the representation on axes
From the graph, draw a horizontal perpendicular line and vertical perpendicular lines to the y-axis and x-axis respectively from point A which looks like
On calculating we find that the value of the x-axis is 0 as the point E lies on the y-axis so only the y-axis value is −4 as the value is down the origin. So the x-axis becomes 0 and the y-axis becomes −4
The ordered pair of E is (0,−4)
Page 95 Exercise 5 Answer
Given: The graph on the points already plotted on that
To find the point lies in Quadrant IV
Solution: the plotted points on the graph is from the cartesian plane
From the given graph, we can clearly see that the points in which values of the x-axis are positive and the y-axis are negative are highlighted in the below graph
From the above graph, we can see that the only point where the x-axis is positive and the y-axis is negative is D. The value of D is(2,−3) which is in quadrant IV
From the graph, the point that lies in Quadrant IV is D.
Page 95 Exercise 6 Answer
Given: A well-plotted graph with different points
To find points located on the negative x-axis
Solution: From the graph, the label of the axes on the negative side
We have the graph where we have to check the left side of origin which shows the negative value of x-axis as
From the above graph, we can clearly see that only one point G is on the left side of the origin and the value of that point is −3 on the x-axis and 0 on the y-axis. So the value of the axis is(−3,0)
The point located on the negative axis is G as shown in the graph
Page 95 Exercise 7 Answer
Given: The equation x−y=−12
To find the solution for y
Solution: Using the given equation, we have to solve it by having y on the left and the remaining values on the right side of the equal sign.
We have the given equation
x−y=−12
Now taking x on the right side, the equation becomes
−y=−12−x
Now taking negative signs from both sides to make them positive, so
−1(y)=−1(12+x)
y=12+x
The solved equation x−y=−12 for y is
y=12+x
Page 95 Exercise 8 Answer
Given: The given equation is 8x+4y=16
To find the solution for y
Solution: From the given equation, we have to solve them by taking y on left and others on the right
Now from the given equation
8x+4y=16
4(2x+y=4)
Taking 4 commons from the whole equation to eliminate
Now we have
2x+y=4
⇒ y=4−2x
Solving 8x+4y=16 for y we get
y=4−2x
Page 95 Exercise 9 Answer
Given: Equation with variables 3x−5y+15=0
To find the solution of y
Solution: we have to use the given equation to solve it for y
We have the equation
3x−5y+15=0
Taking 5y on another side, we get,
3x+15=5y or 5y=3x+15
Now, divide the whole equation with 5 we get
\(\frac{1}{5}\) (5y=3x+15)
⇒y\(=\frac{3}{5} x+3\)
The equation 3x−5y+15=0 solved for y is
y\(=\frac{3}{5} x+3\)
Page 95 Exercise 10 Answer
Given: 0=3y−6x+12
To find, Solve the given equation.
Here we evaluate the equation for y.
By evaluating
3y−6x+12=0
3y=6x−12
3y=3(2x−4)
y\(=\frac{3}{3}(2 x-4)\)
y=2x−4
The value of the given equation is y=2x−4
Page 95 Exercise 11 Answer
Given: y−2=3x+4y
To find, Solve the given equation.
Here we evaluate the equation for y.
By solving
y−2=3x+4y
4y−y=−2−3x
3y=−2−3x
y\(=\frac{-2}{3}-\frac{3 x}{3}\)
y\(=\frac{-2}{3}-x\)
The value of the given equation is y\(=\frac{-2}{3}-x\)