QUESTION IMAGE
Question
summary
- substitution method for solving systems of equations: pick one of the two given equations and define one variable in terms of the other; substitute the expression for the defined variable into the other equation and solve; and substitute this solution back into either equation and solve for the remaining variable.
- combination method for solving systems of equations: eliminate one variable by combining the equations after multiplying either equation by a coefficient if necessary. after solving for one variable, substitute this value into either equation to solve for the remaining variable if required.
exercise 2
directions: choose the best answer from the four choices given. answers are on page 201.
- if $x + y = 4$ and $y - x = 2$, then $x = $?
a. 0
b. 1
c. 2
d. 3
- if $7x = 2$ and $3y - 7x = 10$, then $y = $?
a. 3
b. 4
c. 5
d. 6
- if $\frac{x + y}{2} = 4$ and $x - y = 4$, then $x = $?
a. 2
b. 4
c. 6
d. 8
- if $4x + 3y = 6$ and $3x + 2y = 5$, then $y = $?
a. -6
b. -2
c. 3
d. 6
- if $x + y + z = 10$, $x - y - 2z = -13$, and $x + z = 6$, then $x = $?
a. 1
b. 3
c. 4
d. 5
Question 1
Step1: Define y from second equation
From \( y - x = 2 \), we get \( y = x + 2 \).
Step2: Substitute y into first equation
Substitute \( y = x + 2 \) into \( x + y = 4 \):
\( x + (x + 2) = 4 \)
\( 2x + 2 = 4 \)
\( 2x = 2 \)
\( x = 1 \)
Step1: Find x from first equation
From \( 7x = 2 \), \( x = \frac{2}{7} \).
Step2: Substitute x into second equation
Substitute \( x = \frac{2}{7} \) into \( 3y - 7x = 10 \):
\( 3y - 7(\frac{2}{7}) = 10 \)
\( 3y - 2 = 10 \)
\( 3y = 12 \)
\( y = 4 \)
Step1: Simplify first equation
From \( \frac{x + y}{2} = 4 \), multiply both sides by 2: \( x + y = 8 \).
Step2: Solve system with second equation
We have \( x + y = 8 \) and \( x - y = 4 \).
Add the two equations: \( (x + y) + (x - y) = 8 + 4 \)
\( 2x = 12 \)
\( x = 6 \)
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
B. 1