Expressions and Equations: Analyzing and Solving Systems of Linear Equations (CCSS.8.EE.8)
Common Core 8th Grade Math · Learn by Concept
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Common Core 8th Grade Math › Expressions and Equations: Analyzing and Solving Systems of Linear Equations (CCSS.8.EE.8)
What is the solution to this system? $y = 2x + 1$ and $4x - y = 3$
(2, 5)
(1, 3)
(2, 3)
(-2, -5)
Explanation
Substitute $y=2x+1$ into $4x - y = 3$: $4x - (2x+1)=3 \Rightarrow 2x - 1 = 3 \Rightarrow 2x=4 \Rightarrow x=2$. Then $y=2(2)+1=5$. A common error is dropping the negative when substituting, which leads to $(1,3)$.
Solve the system of linear equations: $2x + y = 6$ and $x + y = 5$
(2, 3)
(1, 4)
(3, 2)
no solution
Explanation
Eliminate $y$ by subtracting: $(2x+y)-(x+y)=6-5 \Rightarrow x=1$. Substitute into $x+y=5$: $1+y=5 \Rightarrow y=4$. Parallel lines would give no solution only if the left sides were identical but constants different.
What ordered pair satisfies both equations? $y = x + 2$ and $2x + y = 8$
(2, 0)
(4, 2)
(2, 4)
(3, 5)
Explanation
Use substitution: $y=x+2$. Then $2x + (x+2) = 8 \Rightarrow 3x+2=8 \Rightarrow 3x=6 \Rightarrow x=2$. So $y=2+2=4$. Errors like using $y=x-2$ lead to wrong points such as $(2,0)$.
Find the intersection point of the lines given by $x + y = 1$ and $x - y = 3$
(1, 2)
(2, 1)
no solution
(2, -1)
Explanation
Add the equations to eliminate $y$: $(x+y)+(x-y)=1+3 \Rightarrow 2x=4 \Rightarrow x=2$. Substitute into $x+y=1$: $2+y=1 \Rightarrow y=-1$. Lines with identical slopes and different intercepts would be parallel with no solution; that is not the case here.
What is the solution to this system? $; y = 2x + 1$ and $; x + y = 7$
$(2, 3)$
$(2, 5)$
$(3, 2)$
No solution
Explanation
Substitute $y=2x+1$ into $x+y=7$: $x+(2x+1)=7 \Rightarrow 3x=6 \Rightarrow x=2$. Then $y=2(2)+1=5$. So $(x,y)=(2,5)$. Other options come from arithmetic slips or thinking the lines are parallel (they are not).
Solve the system by elimination: $; 3x + 2y = 11$ and $; -3x + 2y = 5$
$(-1, 4)$
$(1, 2)$
No solution
$(1, 4)$
Explanation
Add the equations: $(3x+2y)+(-3x+2y)=11+5 \Rightarrow 4y=16 \Rightarrow y=4$. Substitute into $3x+2y=11$: $3x+8=11 \Rightarrow 3x=3 \Rightarrow x=1$. Thus $(1,4)$. Distractors reflect sign mistakes, plugging $y=2$, or misreading as parallel lines.
What is the solution to this system? $; y = -x + 6$ and $; 2x + y = 8$
$(2, 4)$
$(4, 2)$
No solution
$(2, -4)$
Explanation
Substitute $y=-x+6$ into $2x+y=8$: $2x+(-x+6)=8 \Rightarrow x+6=8 \Rightarrow x=2$. Then $y=-2+6=4$. So $(2,4)$. Other choices swap coordinates, flip a sign, or claim parallel lines (not the case).
A club sold adult tickets for 5 dollars and child tickets for 3 dollars. They sold 12 tickets for a total of 48 dollars. Let $x$ be adult tickets and $y$ be child tickets. Solve the system $x + y = 12$ and $5x + 3y = 48$. What is $(x,y)$?
$(8, 4)$
$(4, 8)$
$(6, 6)$
No solution
Explanation
From $x+y=12$, take $x=12-y$. Substitute in $5x+3y=48$: $5(12-y)+3y=48 \Rightarrow 60-5y+3y=48 \Rightarrow -2y=-12 \Rightarrow y=6$. Then $x=12-6=6$. So $(6,6)$. Other pairs come from swapping or miscalculating the substitution.
What is the solution to this system? $y = 2x + 1$ $x + y = 7$
(2, 5)
(3, 4)
(2, 3)
No solution
Explanation
Substitute $y = 2x + 1$ into $x + y = 7$: $x + (2x + 1) = 7 \Rightarrow 3x + 1 = 7 \Rightarrow 3x = 6 \Rightarrow x = 2$. Then $y = 2(2) + 1 = 5$. The lines intersect at $(2, 5)$.
What ordered pair $(x, y)$ satisfies this system? $2x + 3y = 13$ $4x + 3y = 17$
(3, 2)
(2, 3)
(1, 5)
No solution
Explanation
Eliminate $y$ by subtracting: $(4x + 3y) - (2x + 3y) = 17 - 13 \Rightarrow 2x = 4 \Rightarrow x = 2$. Substitute into $2x + 3y = 13$: $4 + 3y = 13 \Rightarrow 3y = 9 \Rightarrow y = 3$. So $(2, 3)$.